HDUoj Virtual Friends （并查集

最新推荐文章于 2018-08-11 20:13:00 发布

原创最新推荐文章于 2018-08-11 20:13:00 发布 · 728 阅读

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online judge HDU 同时被 2 个专栏收录

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数据结构并查集

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本文介绍了一个基于虚拟好友的社交网络问题。通过建立并查集的数据结构来跟踪每个人的社会网络规模，当新的友谊关系形成时，能够实时计算并输出最新形成的社交网络中的人数。

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Virtual Friends

Description

These days, you can do all sorts of things online. For example, you can use various websites to make virtual friends. For some people, growing their social network (their friends, their friends’ friends, their friends’ friends’ friends, and so on), has become an addictive hobby. Just as some people collect stamps, other people collect virtual friends.

Your task is to observe the interactions on such a website and keep track of the size of each person’s network.

Assume that every friendship is mutual. If Fred is Barney’s friend, then Barney is also Fred’s friend.

Input

Input file contains multiple test cases.
The first line of each case indicates the number of test friendship nest.
each friendship nest begins with a line containing an integer F, the number of friendships formed in this frindship nest, which is no more than 100 000. Each of the following F lines contains the names of two people who have just become friends, separated by a space. A name is a string of 1 to 20 letters (uppercase or lowercase).

Output

Whenever a friendship is formed, print a line containing one integer, the number of people in the social network of the two people who have just become friends.

Sample Input

1
3
Fred Barney
Barney Betty
Betty Wilma

Sample Output

2
3
4

题意

给出n组关系，每组关系有两个人的名字，表示这两个人是朋友。 遵循原则：你的朋友的朋友也是你的朋友。需要你在输入每组关系的同时  输出在这对关系的前提下有多少个人刚刚成为朋友。

题解：

并查集板子遇到了玄学QAQ

AC代码

#include <bits/stdc++.h>
using namespace std;
const int N = 100010*2;
int par[N], num[N];
map<string,int> mp;
void init()
{
    for(int i = 1;i < N; i++) {
        par[i] = i; num[i] = 1;
    } 
}
int find(int x) {
    if(x == par[x]) return x;
    return par[x] = find(par[x]);
} 
void unite(int x,int y)
{
    x = find(x);
    y = find(y);
    if(x != y) {
        par[x] = y;
        num[y] += num[x];
    }
}
int main()
{
    int T;
    char a[100], b[100];
    while(~scanf("%d",&T)) {
        while(T--) {
            mp.clear();
            int u, v;
            int n;
            scanf("%d",&n);
            init();
            int ans = 1;
            for(int i = 0;i < n; i++) { 
                scanf("%s %s",a,b);

                if(!mp[a]) mp[a] = ans++;
                if(!mp[b]) mp[b] = ans++;
                unite(mp[a],mp[b]);
                printf("%d\n",num[find(mp[b])]);
            }   
        }
    }
return 0;
}