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夭辰2023-10-30 22:34

CQUPT 的某数据结构homework

基于线性表的图书信息管理

首先，因为我们要输入输出汉语，所以需要更改编码成为utf-8

我使用的是devcpp

所以其他版本编译器更改编码的就请自己去查询更改方式。

工具：

编译选项：

编译时加入以下命令，打勾，下面粘上此代码

-std=c++11 -finput-charset=utf-8

cpp 复制代码 
/*
*/
#include<bits/stdc++.h>
using namespace std;
#define ll long long
typedef  pair<int,int> pi ;
#define if1(x) for(int i =1 ;i<=x;i++)
#define if0(x) for(int i = 0;i<x;i++)
#define jf0(x) for(int j = 0;j<x;j++)
#define jf1(x) for(int j = 1;j<=x;j++)
struct node_book
{
    string name_book;
    string  id_book;
    float price_book;
    //int myid;

};
unordered_map<string ,int>mmp;//去重用
list<node_book>  books;
vector<node_book>  high_books;
double evrage;
int books_num,idx;
void output(){
    for(auto j:books){
        cout<<j.id_book<<" "<<j.name_book<<" ";
        printf("%.2f\n",j.price_book);
    }
}
void add(string id,float price, string name){
    node_book new_book;
    new_book.id_book = id;
    new_book.name_book = name;
    new_book.price_book = price;
    books.push_back(new_book);
    books_num++;
    evrage+= price;
}
void init(){
    books_num = 0;
    float price;
    string name,id;
    while(1){
        cin>>id>>name>>price;
        if(id=="0"&&name=="0"&&price==0)break;
        add(id,price,name);
        mmp[id]++;
       // evrage += price;
    }
}
void func1(){
    init();
    cout<<books_num<<endl;
    output();
}
void func2(){
    init();
    evrage /= books_num;
    for(auto& j:books){
        float te = j.price_book;
        if(te < evrage)  j.price_book*=1.2;
        else   j.price_book*=1.1;
        
    }
    printf("%.2f\n",evrage);
    cout<<books_num<<endl;
    output();
}
void func3(){
    init();
    high_books.clear();
    high_books.push_back(*books.begin());
    float high_price = high_books[0].price_book;
    for(auto j:books){
        if(j.price_book>high_price){
            high_books.clear();
            high_books.push_back(j);
            high_price = j.price_book;

        }
        else if(high_price==j.price_book){
            high_books.push_back(j);
        }
    }
    cout<<high_books.size()<<endl;
    for(auto j: high_books){
        cout<<j.id_book<<" "<<j.name_book<<" ";
        printf("%.02f\n",j.price_book);
    }
}
void func4(){
    init();
    float price;
    string name,id;
    int place,idx = 1;
    cin>>place;
    cin>>id>>name>>price;
    node_book new_book;
    new_book.id_book = id;
    new_book.name_book = name;
    new_book.price_book = price;
    list<node_book>::iterator pos ;
    pos = books.begin();
    for(auto j:books){
        idx++;
        pos++;
        if(idx==place){
        books.insert(pos,new_book);
        books_num++;
    	cout<<books_num<<endl;
    	output();
    	return;
        }
    }
    cout<<"抱歉，入库位置非法!\n";
    
}
void func5(){
    init();
    float price;
    string name,id;
    int place,idx = 1;
    cin>>place;
    list<node_book>::iterator pos ;
    pos = books.begin();
    for(auto j:books){
        idx++;
        pos++;
        if(idx==place){
        books.erase(pos);
        books_num--;
        }
    }
    cout<<books_num<<endl;
    output();
}
void func6(){
    init();
    int place;
    list<node_book>::iterator pos ;
    vector<list<node_book>::iterator> dela;
    pos = books.begin();
    for(auto j:books){
        //cout<<mmp[j.id_book]<<endl;
        if(mmp[j.id_book]>1){
            mmp[j.id_book]--;
            dela.push_back(pos);
            books_num--;
        }
        pos++;
    }
    for(auto it:dela){
        books.erase(it);
    }
    cout<<books_num<<endl;
    output();
}
void tishi(){
   printf("\n输入1测试：基于顺序（链式）存储结构的图书信息表的创建和输出\n");
   cout<<"输入2测试：基于顺序（链式）存储结构的图书信息表的修改\n";
   cout<<"输入3测试：基于顺序（链式）存储结构的图书信息表的最贵图书的查找\n";
   cout<<"输入4测试：基于顺序（链式）存储结构的图书信息表的新图书的入库\n";
   cout<<"输入5测试：基于顺序（链式）存储结构的图书信息表的旧图书的出库\n";
   cout<<"输入6测试：基于顺序（链式）存储结构的图书信息表的图书去重\n";
   cout<<"输入-1，退出程序\n";
}
int main(){
    int op =1;
    while(op!=0){
    tishi();
    cin>>op;
    if(op== 1)
    func1();
    else if(op==2)
    func2();
    else if(op==3)
     func3();
    else if(op==4)
    func4();
    else if(op==5)
    func5();
    else if(op==6)
    func6();
    
    books.clear();
    
    }
    return 0;
}

基于栈的算术表达式求值

普通应用即可

cpp 复制代码 
#include<bits/stdc++.h>
using namespace std;
#define MAXSIZE 100
#define N 10
string S;
//运算符栈 
unordered_map<char,int>mmp;
void init0(){
    mmp['('] = mmp[')']=3;
    mmp['*'] = mmp['/'] = 2;
    mmp['+'] = mmp['-'] = 1;
}

stack<char> sta_op;
stack<int> sta_num;
int verify(char ch)
{
    //cout<<ch<<endl;
    if(ch=='+')
        return 0;
    else if(ch=='-')
        return 1;
    else if(ch=='*')
        return 2;
    else if(ch=='/')
        return 3;
    else if(ch=='(')
        return 4;
    else if(ch==')')
        return 5;
    else
    {
        return 100;
    }
        
}
void push_num(int& idx){
            int j = idx;
            int temp = 0;
            while(S[j]>='0'&&S[j]<='9'){
                temp*=10;
                temp+=S[j]-'0';
                j++;
            }
            sta_num.push(temp);
            idx = j-1;
}
int judge(int a,char e,int b)
{
    //cout<<e<<endl;
    if(e=='+')
        return a+b;
    else if(e=='-')
        return a-b;
    else if(e=='*')
        return a*b;
    else if(e=='/')
        return a/b;
    else
    {
        cout<<"error!"<<endl;
        return -1;
    }
        
}
void func1(int& idx){
    char ch = S[idx];
    if(ch=='('||ch=='+'||ch=='-')
     sta_op.push(ch);//of course +-不需要考虑运算顺序
    else if(ch == '*'||ch=='/'){
        if(S[idx+1]>='0'&&S[idx+1]<='9'){
            //*/在后面跟上的是数字的情况下，一定是可以直接进行运算的
            //那么不妨在这种情况下面直接开算
            int j = idx+1;
            int temp = 0;
            while(S[j]>='0'&&S[j]<='9'){
                temp*=10;
                temp+=S[j]-'0';
                j++;
            }
            sta_num.push(temp);
            idx = j-1;

            //

            int la = sta_num.top();
            sta_num.pop();
            int pre = sta_num.top();
            sta_num.pop();
            int ans = judge(pre,ch,la);
            sta_num.push(ans);
        }
        else{
            sta_op.push(ch);//只能等后面在来运算辣
            //就是等下一个左括号运算完成并出栈的时候来运算此符号
        }
    }else if(ch==')'){
        while(sta_op.top()!='('){
            int la = sta_num.top();
            sta_num.pop();
            int pre = sta_num.top();
            sta_num.pop();
            int ans = judge(pre,sta_op.top(),la);
            sta_op.pop();
            sta_num.push(ans);
        }
        sta_op.pop();
        if(sta_op.top()=='*'||sta_op.top()=='/'){//
        //这个时候就要运算89行处的*/法辣
            int la = sta_num.top();
            sta_num.pop();
            int pre = sta_num.top();
            sta_num.pop();
            int ans = judge(pre,sta_op.top(),la);
            sta_op.pop();
            sta_num.push(ans);
        }
    }

}
int func(){
    cin>>S;
    if(S=="=")return 0;
    for(int i=0;S[i]!='\0';i++)
    {
        if(S[i]>='0'&&S[i]<='9')
            push_num(i);
        else
            func1(i);
     }
     //*/法已经算完辣，+-法没有顺序可言，但是为了减少代码量就不优化辣。
     while(!sta_op.empty()){
            int la = sta_num.top();
            sta_num.pop();
            int pre = sta_num.top();
            sta_num.pop();
            int ans = judge(pre,sta_op.top(),la);
            sta_op.pop();
            sta_num.push(ans);
     } 
    cout<<sta_num.top()<<endl;
    sta_num.pop();
    return 1;
}
int main()
{
    int ti=1;
    while(ti){
        cout<<"\n请输入中缀表达式的算数表达式\n仅输入一个 = 时结束程序\n";
        if(!func())break;

    }
    return 0;
}

基于字符串模式匹配算法的病毒感染检测问题

kmp即可

cpp 复制代码 
#include<bits/stdc++.h>
using namespace std;
#define ll long long

typedef  pair<int,int> pi ;
#define if1(x) for(int i =1 ;i<=x;i++)
#define if0(x) for(int i = 0;i<x;i++)
#define jf0(x) for(int j = 0;j<x;j++)
#define jf1(x) for(int j = 1;j<=x;j++)
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;
const int N = 1e5 + 10, M = 1e6 + 10;
char p[N], s[M];
int n, m;
int ne[N];

void tishi(){
    cout<<"请输入病毒的ONA序列和人的NDNA序列,用空格隔开\n";
    cout<<"当输入的两个DNA序列为 0 0 时退出程序！\n";
}
bool solve(){
    string x,y;
    cin>>x>>y;
    n = y.size();
    m = x.size();
    if0(n){
        s[i+1]= y[i];
    }
    if0(m) p[i+1] = x[i];
    if(x=="0"&&y=="0")return 0;

    memset(ne,0,sizeof(ne));
    for (int i = 2, j = 0; i <= m; i ++ ){
    while (j && p[i] != p[j + 1]) j = ne[j];
    if (p[i] == p[j + 1]) j ++ ;
    ne[i] = j;
}

// 匹配
for (int i = 0, j = -1; i <= n; i ++ ){
    while (j && s[i] != p[j + 1]) j = ne[j];
    if (s[i] == p[j + 1]) j ++ ;
    if(j==m){
        cout<<"YES\n";
        return 1;
    }
    if (j == m)
    {
        j = ne[j];
        // 匹配成功后的逻辑
    }
}
cout<<"NO\n";
return 1;

}
int main(){
    int t=1;
    while (t)
    {
        tishi();
        if(!solve()) {
            break;
        }
        
    }
   // getchar();
    
    return 0;
}

基于哈夫曼树的数据压缩算法

这个还是比较有趣，可以自己试试

cpp 复制代码 
/*
*/
#include<bits/stdc++.h>
using namespace std;
#define ll long long

typedef  pair<int,int> pi ;
#define if1(x) for(int i =1 ;i<=x;i++)
#define if0(x) for(int i = 0;i<x;i++)
#define jf0(x) for(int j = 0;j<x;j++)
#define jf1(x) for(int j = 1;j<=x;j++)
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;
string s;
string ans[1000];
unordered_map<char,int> mmp;
vector<tuple<int,int,int,int>>tre;//重，fa，l,r
bool cmp(pair<char,int> a,pair<char,int> b)
{
    return a.first>b.first;
}
void dfs(int x,string temp){
        if(temp!="\0")
        ans[x] =temp ;
        int w1,w2,f1,f2,l1,l2,r1,r2,id1,id2;
        auto te1 = tre[x];
        tie(w1,f1,l1,r1) = te1;
        if(l1!=-1)dfs(l1,temp+'0');
        if(r1!=-1)dfs(r1,temp+'1');
    
}
void solve(){
    int m,n,idx = 0;
    tre.clear();
    memset(ans,'\0',sizeof(ans));
    priority_queue<pi,vector<pi>,greater<pi>>mq;
    mmp.clear();
    for(auto j:s)mmp[j]++;
    vector<pair<int,int> > ss;
    for(auto j:mmp){
        ss.push_back({j.first,j.second});
    }
    sort(ss.begin(),ss.end());


    for(auto j:ss){
        char ch = j.first;
        printf("%c:%d ",ch,j.second);
        mq.push({j.second,idx++});
        //建树
        tre.push_back({j.second,-1,-1,-1});
    }
    cout<<endl;
    //建立合成节点的树
    while(mq.size()>1){
        auto t1 = mq.top();mq.pop();
        auto t2 = mq.top();mq.pop();
        int w1,w2,f1,f2,l1,l2,r1,r2,id1,id2;
        tie(w1,id1) = t1;tie(w2,id2)  =t2;
        mq.push({w1+w2,idx++});
        tre.push_back({w1+w2,-1,id1,id2});
        //以后要牢记tuple一般用于无修改的操作
        //不是说内容一旦过长就用tuple；
        auto te1 = tre[id1],te2 = tre[id2];
        tie(w1,f1,l1,r1) = te1;
        tie(w2,f2,l2,r2) = te2;
        tre[id1] = {w1,idx-1,l1,r1};
        tre[id2] = {w2,idx-1,l2,r2};

    }

    //输出树
    jf0(idx){
        int w1,w2,f1,f2,l1,l2,r1,r2,id1,id2;
        auto te1 = tre[j];
        tie(w1,f1,l1,r1) = te1;
        cout<<j+1<<" "<<w1<<" "<<f1+1<<" "<<l1+1<<" "<<r1+1<<endl;
    }

    //dfs编码树
    dfs(idx-1,"\0");
    if0(ss.size()){
        char ch = ss[i].first;
        cout<<ch<<":"<<ans[i]<<" ";
    }
    cout<<endl;
    string res;
    if0(ss.size()){
        jf0(ss[i].second){
            res+=ans[i];
        }
    }
    cout<<res<<endl;

    //解码，感觉会比较复杂
   //奥！，dfs编码，那么我也应该是dfs解码，聪明like me
   int now = idx-1;
   for(auto j:res){
        int w1,w2,f1,f2,l1,l2,r1,r2,id1,id2;
        auto te1 = tre[now];
        tie(w1,f1,l1,r1) = te1;
        if(l1==-1){//走到尽头辣,我们就可以输出当前节点的val，并且回溯到根节点
        char ch = ss[now].first;
            cout<<ch;now = idx-1;
            auto te1 = tre[now];
            tie(w1,f1,l1,r1) = te1;
        }
    if(j=='0'){
        //向左边走、
        now = l1;
    }
    else now = r1;
   }
    char ch = ss[now].first;
    cout<<ch;now = idx-1;
    cout<<endl;
}
int main(){
    //ios::sync_with_stdio(false);
    //cin.tie(nullptr);
    //cout.tie(nullptr); 
    
    while (1)
    {
        cin>>s;
        if(s=="0")break;
        solve();
    }
   // getchar();
    
    return 0;
}```

## 如何改变文本的样式

*强调文本* _强调文本_

**加粗文本** __加粗文本__

==标记文本==

~~删除文本~~

> 引用文本

H~2~O is是液体。

2^10^ 运算结果是 1024.

## 插入链接与图片

链接: [link](https://www.csdn.net/).

图片: ![Alt](https://imgconvert.csdnimg.cn/aHR0cHM6Ly9hdmF0YXIuY3Nkbi5uZXQvNy83L0IvMV9yYWxmX2h4MTYzY29tLmpwZw)

带尺寸的图片: ![Alt](https://imgconvert.csdnimg.cn/aHR0cHM6Ly9hdmF0YXIuY3Nkbi5uZXQvNy83L0IvMV9yYWxmX2h4MTYzY29tLmpwZw =30x30)

居中的图片: ![Alt](https://imgconvert.csdnimg.cn/aHR0cHM6Ly9hdmF0YXIuY3Nkbi5uZXQvNy83L0IvMV9yYWxmX2h4MTYzY29tLmpwZw#pic_center)

居中并且带尺寸的图片: ![Alt](https://imgconvert.csdnimg.cn/aHR0cHM6Ly9hdmF0YXIuY3Nkbi5uZXQvNy83L0IvMV9yYWxmX2h4MTYzY29tLmpwZw#pic_center =30x30)

当然，我们为了让用户更加便捷，我们增加了图片拖拽功能。

## 如何插入一段漂亮的代码片

去[博客设置](https://mp.csdn.net/console/configBlog)页面，选择一款你喜欢的代码片高亮样式，下面展示同样高亮的 `代码片`.
```javascript
// An highlighted block
var foo = 'bar';

基于二叉树的表达式求值算法

最傻逼的作业

cpp 复制代码 
#include<bits/stdc++.h>
using namespace std;
#define MAXSIZE 100
#define N 10
string S;
//运算符栈 
unordered_map<char,int>mmp;
void init0(){
    mmp['('] = mmp[')']=3;
    mmp['*'] = mmp['/'] = 2;
    mmp['+'] = mmp['-'] = 1;
}
struct node{
    bool isnum;
    int val;
    char op;
    node* L;
    node* R;
};
stack<char> sta_op;
stack<node*> sta_num;
int verify(char ch)
{
    //cout<<ch<<endl;
    if(ch=='+')
        return 0;
    else if(ch=='-')
        return 1;
    else if(ch=='*')
        return 2;
    else if(ch=='/')
        return 3;
    else if(ch=='(')
        return 4;
    else if(ch==')')
        return 5;
    else
    {
        return 100;
    }
        
}
void push_num(int& idx){
            int j = idx;
            int temp = 0;
            while(S[j]>='0'&&S[j]<='9'){
                temp*=10;
                temp+=S[j]-'0';
                j++;
            }
            node* te = new(node);
            te->isnum=1;
            te->val=temp;
            sta_num.push(te);
            idx = j-1;
}
int judge(int a,char e,int b)
{
    //cout<<e<<endl;
    if(e=='+')
        return a+b;
    else if(e=='-')
        return a-b;
    else if(e=='*')
        return a*b;
    else if(e=='/')
        return a/b;
    else
    {
        cout<<"error!"<<endl;
        return -1;
    }
        
}
void func1(int& idx){
    char ch = S[idx];
    if(ch=='('||ch=='+'||ch=='-')
     sta_op.push(ch);//of course +-不需要考虑运算顺序
    else if(ch == '*'||ch=='/'){
        if(S[idx+1]>='0'&&S[idx+1]<='9'){
            //*/在后面跟上的是数字的情况下，一定是可以直接进行运算的
            //那么不妨在这种情况下面直接开算
            int j = idx+1;
            int temp = 0;
            while(S[j]>='0'&&S[j]<='9'){
                temp*=10;
                temp+=S[j]-'0';
                j++;
            }
            node* te = new(node);
            te->isnum=1;
            te->val=temp;
            sta_num.push(te);
            idx = j-1;

            //

            node* re = sta_num.top();
            sta_num.pop();
            node* la = sta_num.top();
            sta_num.pop();
            node* fa =new(node);
            fa->isnum=0;
            fa->op = ch;
            fa->L=la;
            fa->R = re;
            sta_num.push(fa);
        }
        else{
            sta_op.push(ch);//只能等后面在来运算辣
            //就是等下一个左括号运算完成并出栈的时候来运算此符号
        }
    }else if(ch==')'){
        while(sta_op.top()!='('){
            node* re = sta_num.top();
            sta_num.pop();
            node* la = sta_num.top();
            sta_num.pop();
            node* fa =new(node);
            fa->isnum=0;
            fa->op = sta_op.top();
            fa->L=la;
            fa->R = re;
            sta_num.push(fa);
            sta_op.pop();
        }
        sta_op.pop();
        if(sta_op.top()=='*'||sta_op.top()=='/'){//
        //这个时候就要运算109行处的*/法辣
            node* re = sta_num.top();
            sta_num.pop();
            node* la = sta_num.top();
            sta_num.pop();
            node* fa =new(node);
            fa->isnum=0;
            fa->op = sta_op.top();
            fa->L=la;
            fa->R = re;
            sta_num.push(fa);
            sta_op.pop();
        }
    }

}
int dfs(node* root){
    if(root->isnum) return root->val;
    int l = dfs(root->L),r=dfs(root->R);
     return judge(l,root->op,r);
}
int func(){
    cin>>S;
    if(S=="=")return 0;
    for(int i=0;S[i]!='\0';i++)
    {
        if(S[i]>='0'&&S[i]<='9')
            push_num(i);
        else
            func1(i);
     }
     //*/法已经算完辣，+-法没有顺序可言，但是为了减少代码量就不优化辣。
     while(!sta_op.empty()){
            node* re = sta_num.top();
            sta_num.pop();
            node* la = sta_num.top();
            sta_num.pop();
            node* fa =new(node);
            fa->isnum=0;
            fa->op = sta_op.top();
            fa->L=la;
            fa->R = re;
            sta_num.push(fa);
            sta_op.pop();
     } 
    node* root = sta_num.top();
    cout<<dfs(root)<<endl;
    sta_num.pop();
    return 1;
}
int main()
{
    int ti=1;
    while(ti){
        cout<<"\n请输入中缀表达式的算数表达式\n仅输入一个 = 时结束程序\n";
        if(!func())break;

    }
    return 0;
}

基于 Dijsktra 算法的最短路

cpp 复制代码 
/*
*/
#include<bits/stdc++.h>
using namespace std;
#define ll long long

typedef  pair<int,int> pi ;
#define if1(x) for(int i =1 ;i<=x;i++)
#define if0(x) for(int i = 0;i<x;i++)
#define jf0(x) for(int j = 0;j<x;j++)
#define jf1(x) for(int j = 1;j<=x;j++)
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;
int e[200],h[200],ne[200],w[200],idx,idx1;
unordered_map<string ,int> mmp;
string xiufu[200];
int dist[200];
int pre[200];
int m,n,c;
void add(int a,int b,int c){
    e[idx] = b;
    w[idx] = c;
    ne[idx] = h[a];
    h[a] = idx++;

}

void dijsktra(int x){
    memset(dist,inf,sizeof(dist));
    priority_queue<pi> mq;
    mq.push(make_pair(0,x));
    dist[x]=0;
    while(!mq.empty()){
        int j = mq.top().second;
        int distance = mq.top().first;
        mq.pop();
        for(int p = h[j];~p;p = ne[p]){
            int q = e[p];//q是终点
            if(distance+w[p]<dist[q]){
                dist[q] = distance+w[p];
                mq.push(make_pair(dist[q],q));
                pre[q] = j;
            }
        }
    }
}
void solve(){
    string s;
    string a,b;
    idx1 = 1;
    idx = 0;
    mmp.clear();
    if0(n)cin>>s;
    memset(h,-1,sizeof(h));
    if0(m){
        cin>>a>>b>>c;
        if(mmp[a]==0){
            xiufu[idx1] = a;
            mmp[a] = idx1++;
        }
        if(mmp[b]==0){
            xiufu[idx1] = b;
            mmp[b] = idx1++;
        }
        add(mmp[a],mmp[b],c);
    }
    cin>>a>>b;
    memset(pre,-1,sizeof(pre));
    dijsktra(mmp[a]);
    stack<string>wedg;
    int en = mmp[b];
    while(en!=-1){
        wedg.push(xiufu[en]);
        en = pre[en];
    }
    cout<<dist[mmp[b]]<<endl;
    while(!wedg.empty()){
        cout<<wedg.top()<<" ";
        wedg.pop();
    }
    cout<<endl;
    
}
int main(){
    //ios::sync_with_stdio(false);
    //cin.tie(nullptr);
    //cout.tie(nullptr); 
    while(1){
        cin>>n>>m;
        if(m==0&&n==0)break;
        solve();
    }
    
    return 0;
}

基于广度优先搜索的六度空间

cpp 复制代码 
/*
*/
#include<bits/stdc++.h>
using namespace std;
#define ll long long

typedef  pair<int,int> pi ;
#define if1(x) for(int i =1 ;i<=x;i++)
#define if0(x) for(int i = 0;i<x;i++)
#define jf0(x) for(int j = 0;j<x;j++)
#define jf1(x) for(int j = 1;j<=x;j++)
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;
vector<int> wedg[10005];
bool marked[10005];
int frend;
int m,n;
queue<pi>mq;
void bfs(){
    while(!mq.empty()){
        auto j = mq.front();
        mq.pop();
        if(marked[j.first]||j.second>6)continue;
        marked[j.first]=1;
        frend++;
    for(auto t:wedg[j.first]){
        if(!marked[t])mq.push(make_pair(t,j.second+1));
    }
    }


}
void solve(){
    if0(m){
        int a,b;
        cin>>a>>b;
        wedg[a].push_back(b);
        wedg[b].push_back(a);
    }
    if1(n){
        double ans;
        memset(marked,0,sizeof(marked));
        frend = 0;
        mq.push({i,0});
        bfs();
        ans = 1.*(frend)/n;
        printf("%d:%.2f%\n",i,ans*100);

    }
    if1(n)wedg[i].clear();
}
int main(){
    while(1){
    cin>>n>>m;
    if(n==0&&m==0)return 0;
    solve();
    }
   // getchar();
    
    return 0;
}

排序算法的实现与分析

归并法

cpp 复制代码 
/*
*/
#include<bits/stdc++.h>
using namespace std;
#define ll long long

typedef  pair<int,int> pi ;
#define if1(x) for(int i =1 ;i<=x;i++)
#define if0(x) for(int i = 0;i<x;i++)
#define jf0(x) for(int j = 0;j<x;j++)
#define jf1(x) for(int j = 1;j<=x;j++)
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;
struct node{
    string name;
    double grade;
};
bool cmp(node a,node b){
    return a.grade>b.grade;
}
void hebin(vector<node>& da,int l,int r,int mid){
    node* temp = new node[r-l+2];
    int i = l,j = mid+1,k=0;
    while(i<=mid&&j<=r){
        if(cmp(da[i],da[j])){
            temp[k++] = da[i++];
        }else{
            temp[k++] = da[j++];
        }
    }
    while(i<=mid)
    temp[k++] = da[i++];
    while(j<=r) temp[k++] = da[j++];
    k=0;
    for(int i = l; i<=r;)da[i++] = temp[k++];
    delete []temp;
}
void guibin(int l,int r,vector<node>& da){
    if(l<r){
        int mid = l+r>>1;
        guibin(l,mid,da);
        guibin(mid+1,r,da);
        hebin(da,l,r,mid);
    }
}
int main(){
    string s;
    double grade;
    vector<node> date;
    while(1){
        cin>>s>>grade;
        if(s=="0"&&grade==0)break;
        node te ;
        te.grade = grade;
        te.name = s;
        date.push_back(te);
    }
   guibin(0,date.size()-1,date);
    
    if0(date.size()){
        cout<<date[i].name<<" ";
        printf("%.2f\n",date[i].grade);
    }
    return 0;
}
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