目录
题目一：
cpp
#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
int g[15][15], vis[15] = { 0 };
int n, m;
queue<int> q;
void dfs(int k)//深度
{
cout << k << " ";
vis[k] = 1;
for (int i = 0; i < n; i++)
{
if (i == k)
continue;
if (vis[i] == 0 && g[k][i])
{
dfs(i);
}
}
return;
}
void bfs(int k)//广度
{
q.push(k);
while (!q.empty())
{
k = q.front();
q.pop();
if(vis[k]==0)
{
cout<<k<<" ";
vis[k]=1;
}
for (int i = 0; i < n; i++)
{
if (vis[i] == 0 && g[k][i] == 1)
{
q.push(i);
}
}
}
return;
}
int main()
{
cin >> n >> m;
for (int i = 1; i <= m; i++)//构建邻接矩阵
{
int x, y;
cin >> x >> y;
g[x][y] = 1, g[y][x] = 1;
}
for (int i = 0; i < n; i++)//深度
{
if (vis[i] == 0)
{
cout << "{ ";
dfs(i);
cout << "}" << endl;
}
}
memset(vis, 0, sizeof(vis));
for (int i = 0; i < n; i++)//广度
{
if(vis[i]==0)
{
cout << "{ ";
bfs(i);
cout << "}" << endl;
}
}
}题目二：
cpp
#include<iostream>
#include<algorithm>
using namespace std;
int g[10005][10005];
float n, k;
typedef struct node
{
int data;
int w = 0;
}node;
void warshall()//传递闭包
{
for (int k = 1; k <= n; k++)
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
{
if (g[i][k] && g[k][j])//连通
{
if (i == j)
continue;
if (g[i][j] == 0 || g[i][j] > g[i][k] + g[k][j])//没有直接连通或者新通路距离小于之前通路
g[i][j] = g[i][k] + g[k][j];
}
}
}
bool cmp(node a, node b)
{
return a.w < b.w;
}
int main()
{
cin >> n >> k;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
g[i][j] = 0;
for (int i = 0; i < k; i++)//无向图，初始距离都为1
{
int v1, v2;
cin >> v1 >> v2;
g[v1][v2] = 1;
g[v2][v1] = 1;
}
warshall();
/*for (int i = 1; i <= n; i++)//输出邻接矩阵
{
for (int j = 1; j <= n; j++)
cout << g[i][j] << " ";
cout << endl;
}*/
float ans[10005] ;
for (int i = 1; i <= n; i++)
{
ans[i] = 0;
for (int j = 1; j <= n; j++)
{
if (i == j)
ans[i]++;
if (g[i][j]>0 && g[i][j] <= 6)//符合条件
ans[i]++;
}
}
for (int i = 1; i <= n; i++)
printf("%d:% .2f%%\n", i, ans[i] / n * 100);
}题目三：
cpp
#include<iostream>
using namespace std;
int g[1010][1010];
int n,m,s;
int vis[1010];
int path[1010];
int cnt=0;
void dfs(int k)//深度
{
//cout<<k<<" ";
path[cnt++]=k;//记录去的路径
vis[k]=1;
for(int i=0;i<=n;i++)
{
if(i==k)
continue;
if(vis[i]==0&&g[k][i]==1)
{
dfs(i);
path[cnt++]=k;//记录回的路径
}
}
return ;
}
int main()
{
cin>>n>>m>>s;
for(int i=1;i<=m;i++)//构建邻接矩阵
{
int x,y;
cin>>x>>y;
g[x][y]=g[y][x]=1;
}
dfs(s);
int flag=1;//标记是否可以遍历完所有
for(int i=1;i<=n;i++)//查询是否遍历完所有
{
if(vis[i]==0)
{
flag=0;
break;
}
}
if(flag==1)//输出序列
{
for(int i=0;i<cnt;i++)
{
cout<<path[i];
if(i!=cnt-1)
cout<<" ";
}
}
else
{
for(int i=0;i<cnt;i++)
cout<<path[i]<<" ";
cout<<0;
}
}题目四：
cpp
#include<iostream>
#include<queue>
#include<vector>
#include<cstring>
using namespace std;
typedef long long ll;
const ll N = 3e5 + 10, M = 1e6 + 10, inf = 1e14;
struct node
{
ll v, w;
bool operator <(const node& y) const//重载一个<,用于优先队列排序
{
return w > y.w;//小到大
}
};
int n, m, t;
priority_queue<node>q;
vector<node> e[N];
ll dis[N], vis[N];
void Dij(int t)
{
dis[t] = 0;//从t号点出发，表示从t到t距离为0
q.push({ t,dis[t] });//t号点以及到t的距离入队
while (q.size())
{
int u = q.top().v;//取最小边相连的点出队
q.pop();
if (vis[u])//访问过则跳过
continue;
vis[u] = 1;//没访问赋为访问
for (auto i : e[u])//遍历以u为出发点的边
{
int v = i.v, w = i.w;//取其相连的点及权值
if (dis[v] > dis[u] + w)//经过u点到v点的权值小于之前的权值则更新
{
dis[v] = dis[u] + w;
q.push({ v,dis[v] });//v点以及到v的距离
}
}
}
}
int main()
{
cin >> n >> m;
for (int i = 1; i <= m; i++)
{
int x, y, w=1;
cin >> x >> y;
e[x].push_back({ y,w });//存入以x出发到y，权值为w
e[y].push_back({ x,w });//存入以x出发到y，权值为w
}
int N;
cin >> N;
while (N--)
{
cin >> t;
float ans = 0,sum=0;
memset(dis, 0x3f3f3f, sizeof(dis));
memset(vis, 0, sizeof(vis));
Dij(t);//从t点出发
for (int i = 1; i <= n; i++)
{
sum += dis[i];//求和
}
//cout << sum;
ans = (n - 1) /sum;
printf("Cc(%d)=%.2f\n", t, ans);
}
}题目五：
cpp
#include<iostream>//并查集
using namespace std;
int fa[204], mp[102][102];
int find(int x)//查找
{
if(fa[x]==x) return x;
return fa[x]=find(fa[x]);
}
void Merge(int x, int y)//合并
{
int a=find(x), b=find(y);
if(a==b) return;
else fa[a]=fa[b];
}
int main()
{
int n, m, k;
cin>>n>>m>>k;
int a, b, c;
for(int i=1;i<=n;i++)//初始化
fa[i]=i;
for(int i=0; i<m; i++)
{
cin>>a>>b>>c;
mp[a][b]=c,mp[b][a]=c;
if(c==1)//为朋友则合并，因为朋友的朋友也是朋友
Merge(a, b);
}
for(int i=0; i<k; i++)
{
cin>>a>>b;
if(mp[a][b]==1) cout<<"No problem"<<endl;
else if(mp[a][b]==-1)//是敌对关系
{
if(find(a)==find(b)) cout<<"OK but..."<<endl;//但是有相同的朋友，即在一个集合内
else cout<<"No way"<<endl;
}
else cout<<"OK"<<endl;
}
return 0;
}题目六：
cpp
#include<iostream>
using namespace std;
int n, m, mp[520][520], f[520];
void init() //初始化
{
for (int i = 0; i < n; i++)
f[i] = i;
}
int find(int x) //查询
{
if (x == f[x]) return x;
return f[x] = find(f[x]);
}
void merge(int x, int y)//合并
{
int xx = find(x);
int yy = find(y);
if (xx != yy) f[xx] = yy;
return;
}
int sum()//统计连通块个数
{
int cnt = 0;
for (int i = 0; i < n; i++) //统计连通块个数
{
if (i == find(i)) cnt++;
}
return cnt;
}
int main()
{
int a, b, k, x, cnt = 0, cnt2, flag = 0;
cin >> n >> m;
init();
while (m--)
{
cin >> a >> b;
mp[a][b] = 1;
mp[b][a] = 1;
merge(a, b);
}
cin >> k;
cnt = sum();
if (k == n) flag = 1; //要输出Game Over;
while (k--)
{
cin >> x;
init();
cnt2 = 0; //x去掉后的连通区域个数
for (int i = 0; i < n; i++)
mp[x][i] = mp[i][x] = 0;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
if (mp[i][j]) merge(i, j);
}
}
cnt2 = sum();
if (cnt2 < cnt + 2) cout << "City " << x << " is lost." << endl;
else cout << "Red Alert: City " << x << " is lost!" << endl;//1、去掉i点后，一个连通块至少被分为两个，增加了一个
cnt = cnt2; //2、去掉i点后，再次统计cnt2时，就会把去掉的点i也加上，所以要大于cnt+1
}
if (flag) cout << "Game Over." << endl;
}题目七：
cpp
#include<iostream>
#include<queue>
using namespace std;
int n, id, fa, ma, k;
int f[10100], peo[10100], avgs[10100], avga[10100],st[10100];
struct node
{
int a, b;
}e[10100];
struct family
{
int id, peo, avgs, avga;
bool operator<(const family& x)const
{
if (x.peo * avga == peo * x.avga)//人均相等,则升序排序
return id > x.id;
return x.peo * avga < peo * x.avga;//人均多的在前
}
};
int find(int x)//查找
{
if (f[x] == x) return x;
return f[x] = find(f[x]);
}
void meger(int x, int y)//合并
{
x = find(x), y = find(y);
if (x != y)
{
f[max(x, y)] = min(x, y);//小号id的设为父亲
peo[min(x, y)] += peo[max(x, y)];//家庭人数加到小号id
avgs[min(x, y)] += avgs[max(x, y)];//房屋数量加到小号id
avga[min(x, y)] += avga[max(x, y)];//服务面积加到小号面积
}
}
int main()
{
for (int i = 0; i < 10100; i++) f[i] = i, peo[i] = 1;//初始化，每个父亲节点是自己，自己家庭都有一个自己
cin >> n;
int count = 0;
for (int i = 1; i <= n; i++)
{
cin >> id >> fa >> ma >> k;
if (fa != -1) e[count++] = { id,fa };//存关系
if (ma != -1) e[count++] = { id,ma };//存关系
st[id] = 1;//存在这个人
int kid;
for (int j = 1; j <= k; j++)
{
cin >> kid;
e[count++] = { id,kid };//存关系
}
cin >> avgs[id] >> avga[id];
}
for (int i = 0; i < count; i++)//合并家庭
{
int a = e[i].a, b = e[i].b;
st[a] = st[b] = 1;
meger(a, b);//合并二者为一个家庭
}
priority_queue<family>ans;//优先队列
for (int i = 0; i < 10100; i++)
{
if (st[i] && f[i] == i)//这个人存在且父亲节点是自己
ans.push({ i,peo[i],avgs[i],avga[i] });
}
//count << st[8888] << f[8888];
cout << ans.size() << endl;
while (!ans.empty())
{
family t = ans.top();
ans.pop();
printf("%04d %d %.3lf %.3lf\n", t.id, t.peo, (double)t.avgs / t.peo, (double)t.avga / t.peo);
}
}