并查集模版及其示例

1  先通过查找根节点将该连在一起的连在一起

2 连接完之后,统计总共有多少伙

    #include<iostream>
    #include<cstdio>
    using namespace std;
    int tree[50010];

    int find(int x)            //寻找根节点
    {
        if(x == tree[x])
            return x;
        else
            return tree[x] = find(tree[x]);
    }

    void union(int x,int y)  // 将一伙的连在一起
    {
        int a=find(x);
        int b=find(y);
        if(a!=b) tree[b]=a;
        return;
    }

    int main()
    {
        int n；    总共有n个元素
        while(scanf("%d", &n) != EOF)
            {
            int sum = 0;
            for(int i = 0; i < n; i++)

               tree[i]=i;                        //初始化
            for(int i = 0; i < m; i++)
            {
                int x, y;
                cin >> x >> y;
              union(x, y);
            }

        }
    }

1   统计多少与X一伙的

  bool same(int x, int y)      //判断是否是一伙

     {

         return find(x) == find(y);

      }

  for(int i = 0;i < n;i++)         //该连在一起的连在一起后,统计总共有多少伙
             {

               if( same(i, X))

              sum++;

             }

            printf("%d\n",,sum);

2  统计有多少伙

for(int i = 1; i <= n; i++)

{

if(tree[i] == i)

ans++;

}

The Suspects

Time Limit: 1000MS Memory Limit: 20000KTotal Submissions: 26901 Accepted: 13142

Description

Severe acute respiratory syndrome (SARS), an atypical pneumonia of unknown aetiology, was recognized as a global threat in mid-March 2003. To minimize transmission to others, the best strategy is to separate the suspects from others. 
In the Not-Spreading-Your-Sickness University (NSYSU), there are many student groups. Students in the same group intercommunicate with each other frequently, and a student may join several groups. To prevent the possible transmissions of SARS, the NSYSU collects the member lists of all student groups, and makes the following rule in their standard operation procedure (SOP). 
Once a member in a group is a suspect, all members in the group are suspects. 
However, they find that it is not easy to identify all the suspects when a student is recognized as a suspect. Your job is to write a program which finds all the suspects.

Input

The input file contains several cases. Each test case begins with two integers n and m in a line, where n is the number of students, and m is the number of groups. You may assume that 0 < n <= 30000 and 0 <= m <= 500. Every student is numbered by a unique integer between 0 and n−1, and initially student 0 is recognized as a suspect in all the cases. This line is followed by m member lists of the groups, one line per group. Each line begins with an integer k by itself representing the number of members in the group. Following the number of members, there are k integers representing the students in this group. All the integers in a line are separated by at least one space. 
A case with n = 0 and m = 0 indicates the end of the input, and need not be processed.

Output

For each case, output the number of suspects in one line.

Sample Input

100 42 1 25 10 13 11 12 142 0 12 99 2200 21 55 1 2 3 4 51 00 0

Sample Output

411

题意：有N个人分成M组，M，N下面共有M列，N个人编号从0到N-1。M列每列第一个数k代表该列有k个数，编号为0的称为可能患有病的人为嫌疑犯，只要一组里面有一个嫌疑犯，那么全组都是。现在为了预防病的传染，需要判断有多少个嫌疑犯。我们已经知道编号为0的人是嫌疑犯 且 一个人可能分属不同的几个组。也就是说：假设有有两个组有同一个嫌疑犯，那么这两个组的人都会变成嫌疑犯。

思路：合并每一组里面的人 + 从头查询， 并查集模版题。#include<bits/stdc++.h>    #define MAXN 30000+10    using namespace std;    int tree[MAXN];    int x[MAXN];    int N, M;    void init()    {        for(int i = 0; i < N; i++)            tree[i] = i;    }   int find(int p)    {        if(tree[p] == p)            return p;        else        {           return tree[p] = find(tree[p]);        }    }    void Union(int x, int y)    {        int fx = find(x);        int fy = find(y);        if(fx != fy)            tree[fx] = fy;    }    bool same(int x, int y)    {        return find(x) == find(y);    }    int main()    {        while(scanf("%d%d", &N, &M), N||M)        {            init();            int k;            while(M--)            {                scanf("%d", &k);                for(int i = 0; i < k; i++)                    scanf("%d", &x[i]);                for(int i = 0; i < k-1; i++)                    Union(x[i], x[i+1]);            }            int ans = 0;            for(int i = 0; i < N; i++)            {                if(same(i, 0))                    ans++;            }            printf("%d\n", ans);        }        return 0;    }