The Suspects

題目連結：http://poj.org/problem?id=1611

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

題意：

有m組，0為起點，有0的那一組全是嫌疑人，之後會不斷傳遞到其它組去，問一共有多少人是嫌疑人。

題解：

這題我一開始亂搞了一通，先貼個程式碼。複雜度應該是可以過的。

#include <cstdio>
#include 
 
<cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;

const int N = 30005 ,M = 505;
int n,m;
vector <int > vec[N];
vector <int > g[M];
int vis[N],check[M];

int main(){
    while(scanf("%d%d",&n,&m)){
        if(n==0 && m==0) break;
        queue <int > q;memset(vis,0,sizeof(vis));memset(check,0,sizeof(check));
        for(int i=0;i<=n;i++) vec[i].clear();
        for(int i=0;i<=m;i++) g[i].clear();
        for(int i=1,k;i<=m;i++){
            scanf("%d",&k);
            for(int j=1,num;j<=k;j++){
                scanf("%d",&num);
                vec[num].push_back(i);
                if(!num){
                    if(!check[i]) q.push(i);
                    check[i]=1;
                }
                g[i].push_back(num);
            }
        }
        vis[0]=1;
        while(!q.empty()){
            int now = q.front();q.pop();
            for(int i=0;i<g[now].size();i++){
                int t = g[now][i];
                if(!vis[t]){
                    vis[t]=1;
                    for(int j=0;j<vec[t].size();j++){
                        if(!check[vec[t][j]]){
                            check[vec[t][j]]=1;
                            q.push(vec[t][j]);
                        }
                    }
                }
            }
        }
        int ans = 0;
        for(int i=0;i<=n;i++) if(vis[i]) ans++;
        printf("%d\n",ans);
    }
    return 0;
}

然後再來說說正解。

由於關係會傳遞，我們想到了並查集（合併集合），一開始的時候將每一組合併到同一個集合中。如果一個人在不同的組，那麼根據並查集，這幾個組都會被放入同一個集合中。

最後看看誰的父親和0的父親一樣就好了。

程式碼如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;

const int N = 30005 ,M = 505;
int n,m;
int f[N+M];

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

void Union(int x,int y){
    int fx=find(x),fy=find(y);
    if(fx==fy) return;
    f[fx]=fy;
}

int main(){
    while(scanf("%d%d",&n,&m)){
        if(!n && !m) break;
        for(int i=0;i<=n+m+1;i++) f[i]=i;
        for(int i=1,k;i<=m;i++){
            scanf("%d",&k);
            for(int j=1,num;j<=k;j++){
                scanf("%d",&num);
                Union(num,n+i);
            }
        }
        int ans = 0,std=find(0);
        for(int i=0;i<=n;i++){
            if(std==find(i)) ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}