HDOJ 3172 Virtual Friends（并查集 + STL）

最新推荐文章于 2018-07-17 21:28:14 发布

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最新推荐文章于 2018-07-17 21:28:14 发布

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分类专栏：数据结构文章标签： HDOJ

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本文介绍了一种使用并查集算法来计算在线社交网络中个人社交圈规模的方法。通过跟踪不同用户之间的友谊关系，该算法能够实时更新并返回任意两个新成为朋友的人所在的社交圈子总人数。

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Virtual Friends

Problem 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

题目大意：现在有人喜欢在网上交友，对于一个人来说，他的朋友，他朋友的朋友，他朋友的朋友的朋友都属于他的社交圈子。你的任务是计算出每个人社交圈子中人数的多少。

解题思路：红果果的并查集~注意使用map将字符串映射到数字，还有就是注意输入格式，要写成while(scanf("%d",&t) != EOF)

代码如下：

#include <algorithm>
#include <cctype>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#define EPS 1e-6
#define INF INT_MAX / 10
#define LL long long
#define MOD 1000003
#define PI acos(-1.0)

const int maxn = 100005;
int par[maxn],rank[maxn],num[maxn];
std::map<std::string,int> mapp;
void init(int n)
{
    for(int i = 1;i <= n;i++){
        par[i] = i;
        rank[i] = 0;
        num[i] = 1;
    }
    mapp.clear();
}

int find(int x)
{
    return par[x] == x ? x : par[x] = find(par[x]);
}

void unite(int x,int y)
{
    x = find(x);
    y = find(y);
    if(x == y){
        printf("%d\n",num[x]);
        return ;
    }

    if(rank[x] < rank[y]){
        par[x] = y;
        num[y] += num[x];
        num[x] = 0;
        printf("%d\n",num[y]);
        return ;
    }
    else{
        par[y] = x;
        num[x] += num[y];
        num[y] = 0;
        if(rank[x] == rank[y])
            rank[x]++;
        printf("%d\n",num[x]);
        return ;
    }

}
int main()
{
    char p[22],q[22];
    int t,n;
    while(scanf("%d",&t) != EOF){
        while(t--){
            scanf("%d",&n);
            init(maxn);
            int cnt = 1;
            for(int i = 1;i <= n;i++){
                scanf("%s %s",p,q);
                if(!mapp[p]){
                    mapp[p] = cnt;
                    cnt++;
                }
                if(!mapp[q]){
                    mapp[q] = cnt;
                    cnt++;
                }
                unite(mapp[p],mapp[q]);
            }
        }
    }
    return 0;
}