《算法笔记》9.6小节 数据结构专题(2)并查集 问题 D: More is better

题目描述

Mr Wang wants some boys to help him with a project. Because the project is rather complex, the more boys come, the better it will be. Of course there are certain requirements.Mr Wang selected a room big enough to hold the boys. The boy who are not been chosen has to leave the room immediately. There are 10000000 boys in the room numbered from 1 to 10000000 at the very beginning. After Mr Wang's selection any two of them who are still in this room should be friends (direct or indirect), or there is only one boy left. Given all the direct friend-pairs, you should decide the best way.

输入

The first line of the input contains an integer n (0 ≤ n ≤ 100 000) - the number of direct friend-pairs. The following n lines each contains a pair of numbers A and B separated by a single space that suggests A and B are direct friends. (A ≠ B, 1 ≤ A, B ≤ 10000000)

输出

The output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep.

样例输入
复制代码
3
1 3
1 5
2 5
4
3 2
3 4
1 6
2 6
样例输出
复制代码
4
5

题目大意: 房间里有编号从1~10000000共10000000人,每次给出n对关系,每对关系表示这两个人被选中了且成为朋友。问最后被选中的最大朋友人数是多少,如果没有人被选中,则留下1个人;如果有多组朋友,输出最大的组有多少人。

分析:并查集的应用。不过这个集合很大,因此要在合并的时候,记录合并后组的人数。

cpp 复制代码
#include<algorithm>
#include <iostream>
#include  <cstdlib>
#include  <cstring>
#include   <string>
#include   <vector>
#include   <cstdio>
#include    <queue>
#include    <stack>
#include    <ctime>
#include    <cmath>
#include      <map>
#include      <set>
#define INF 0xffffffff
#define db1(x) cout<<#x<<"="<<(x)<<endl
#define db2(x,y) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<endl
#define db3(x,y,z) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<endl
#define db4(x,y,z,r) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<endl
#define db5(x,y,z,r,w) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<", "<<#w<<"="<<(w)<<endl
using namespace std;

int father[10000005],cnt[10000005];

int findFather(int x,int father[])
{
    int a=x;
    while(father[x]!=x)x=father[x];
    while(a!=father[a])
    {
        int temp=a;
        a=father[a],father[temp]=x;
    }
    return x;
}

void Union(int a,int b,int father[],int cnt[],int &ans)
{
    int faA=findFather(a,father),faB=findFather(b,father);
//    db4(a,b,faA,faB);
    if(faA!=faB)
    {
        father[faA]=faB;
        cnt[faB]+=cnt[faA];
        ans=max(ans,cnt[faB]);
    }
    return;
}

int main(void)
{
    #ifdef test
    freopen("in.txt","r",stdin);
    //freopen("in.txt","w",stdout);
    clock_t start=clock();
    #endif //test

    int T,n,m;
    while(~scanf("%d",&T))
    {
        int ans=-1;
        if(T==0)
        {
            printf("1\n");continue;
        }
        for(int i=1;i<=10000005;++i)
            father[i]=i,cnt[i]=1;
        for(int i=0;i<T;++i)
        {
            scanf("%d%d",&n,&m);
            Union(n,m,father,cnt,ans);
        }
        printf("%d\n",ans);
    }

    #ifdef test
    clockid_t end=clock();
    double endtime=(double)(end-start)/CLOCKS_PER_SEC;
    printf("\n\n\n\n\n");
    cout<<"Total time:"<<endtime<<"s"<<endl;        //s为单位
    cout<<"Total time:"<<endtime*1000<<"ms"<<endl;    //ms为单位
    #endif //test
    return 0;
}
相关推荐
大怪v21 小时前
前端:人工智能?我也会啊!来个花活,😎😎😎“自动驾驶”整起!
前端·javascript·算法
惯导马工1 天前
【论文导读】ORB-SLAM3:An Accurate Open-Source Library for Visual, Visual-Inertial and
深度学习·算法
骑自行车的码农1 天前
【React用到的一些算法】游标和栈
算法·react.js
博笙困了1 天前
AcWing学习——双指针算法
c++·算法
moonlifesudo1 天前
322:零钱兑换(三种方法)
算法
NAGNIP2 天前
大模型框架性能优化策略:延迟、吞吐量与成本权衡
算法
美团技术团队2 天前
LongCat-Flash:如何使用 SGLang 部署美团 Agentic 模型
人工智能·算法
Fanxt_Ja2 天前
【LeetCode】算法详解#15 ---环形链表II
数据结构·算法·leetcode·链表
侃侃_天下2 天前
最终的信号类
开发语言·c++·算法
茉莉玫瑰花茶2 天前
算法 --- 字符串
算法