POJ 1611 The Suspects (并查集)

The Suspects


Time Limit: 1000MS Memory Limit: 20000K
Total Submissions: 35951 Accepted: 17480



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.



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.



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


Sample Input

100 4
2 1 2
5 10 13 11 12 14
2 0 1
2 99 2
200 2
1 5
5 1 2 3 4 5
1 0
0 0


Sample Output









AC 代码

#include <iostream>
using namespace std;

const int MAXN = 31000; /*结点数目上线*/
int pa[MAXN];    /*p[x]表示x的父节点*/
int rank[MAXN];    /*rank[x]是x的高度的一个上界*/

void make_set(int x)
    pa[x] = x;
    rank[x] = 0;

int find_set(int x)
    if(x != pa[x])
        pa[x] = find_set(pa[x]);
    return pa[x];

void union_set(int x, int y)
    x = find_set(x);
    y = find_set(y);
    if(rank[x] > rank[y])/*让rank比较高的作为父结点*/
        pa[y] = x;
        pa[x] = y;
        if(rank[x] == rank[y])

int main()
    int n,m;
        for(int i=0; i<n; i++)  //初始化每一个学生单独属于一个集合
        for(int i=0; i<m; i++)
            int cou,one,tw;
            for(int j=0; j<cou; j++)
                else union_set(one,tw); //和组的第一个成员合并
        int pa0=find_set(0),ans=1;  //查找0所在的集合
        for(int i=1; i<n; i++)
    return 0;


6 只已被捕捉

  • 山鬼谣 Chrome | 76.0.3809.87 Windows 10/11


    • 千千 Chrome | 75.0.3770.142 Windows 10/11

      可以,加上 rank 只是为了让树更平衡一点,算是个人习惯,也可以不需要(当然可能极端情况下会超时)

      • 山鬼谣 Chrome | 76.0.3809.87 Windows 10/11


        • 千千 Chrome | 76.0.3809.100 Windows 10/11


  • 邱雷 Chrome | 56.0.2924.87 Windows 7

    一下学会了并查集的使用,还顺便学了下 map容器!!!!!

    • 千千 Edge | 15.15058 Windows 10/11

      嗯呢 嗯呢,哈哈