首页 > 代码库 > poj 1611(详解)

poj 1611(详解)

2024-07-15 12:58:29   218人阅读

The Suspects
Time Limit: 1000MS Memory Limit: 20000K
Total Submissions: 22217 Accepted: 10805

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 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

4
1
1

Source

Asia Kaohsiung 2003

AC代码：(代码有详细解释)

#include<iostream>
using namespace std;
int f[30010],num[30010];
int find(int x){
    if(x==f[x]) return f[x];
    f[x]=find(f[x]);
    return f[x];              //返回x的父节点
}
void Union(int x,int y){
    int fx=find(x);
    int fy=find(y);
    if(fx==fy) return ;
    if(num[fx]>num[fy]){        //fx作为父节点的节点数多于fy作为父节点的节点数
        f[fy]=fx;               //fx作为fy的父节点
        num[fx]+=num[fy];       //此时fx作为父节点的节点数就要加上fy作为父节点的节点数
    }
    else{
        f[fx]=fy;
        num[fy]+=num[fx];
    }
}
int main(){
    int n,m;
    while(cin>>n>>m){
        if(n==m&& m==0)
            break;
        int i,j;
        for(i=0;i<n;i++){
            f[i]=i;
            num[i]=1;      //num[i]表示以节点i作为父节点的树共有多少个节点
        }
        int x,y,t;
        while(m--){
            cin>>t;
            cin>>x; x=find(x);
            for(int i=1;i<t;i++){
                cin>>y; y=find(y);
                Union(x,y);
            }
        }
        cout<<num[find(0)]<<endl;           //输出0的父节点的节点数
    }
    return 0;
}



联系
我们