HUST 1103 校赛 邻接表-拓扑排序

Description

N students were invited to attend a party, every student has some friends, only if someone’s all friends attend this party, this one can attend the party(ofcourse if he/she has no friends, he/she also can attend it.), now i give the friendship between these students, you need to tell me whether all of them can attend the party.

Note that the friendship is not mature, for instance, if a is b’s friend, but b is not necessary a’s friend. 

Input

Input starts with an integer T(1 <= T <= 10), denoting the number of test case.

For each test case, first line contains an integer N(1 <= N <= 100000), denoting the number of students.

Next n lines, each lines first contains an integer K, denoting the number of friends belong to student i(indexed from 1). Then following K integers, denoting the K friends.

You can assume that the number of friendship is no more than 100000, and the relation like student A is himself’s friend will not be existed. 

Output

For each test case, if all of the students can attend the party, print Yes, otherwise print No. 

Sample Input

2
3
1 2
1 3
1 1
3
1 2
0
1 1

Sample Output

No
Yes

HINT

For the first case:

Student 1 can attend party only if student 2 attend it.

Student 2 can attend party only if student 3 attend it.

Student 3 can attend party only if student 1 attend it.

So no one can attend the party. 

题意：给定n个人， 第i个人依赖k个人，只有k个人都参加了，那么i才会参加。问是否所有人都能参加。

思路：数据量有些大，可能无法用并查集判是否有回路解。学长给出方法是用拓扑排序，最后将节点数量和n比较判断即可。

拓扑排序大致是查找出度为零的所有节点，压入队列，一个个弹出，同时将该点和临边删除。

用到了Vector邻接表，发现储存图真是好用。

但是我的代码还有点问题，vector并不需要开二维的，一维就够了。

 #include <stdio.h>
 #include <iostream>
 #include <string.h>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <utility>
 #define  MAXX 100010
 using namespace std;
 const int  INF = 0x3f3f3f3f;
 pair<vector<int>, int> x;
 vector< vector<int> >bel();
 queue< vector<int> >qu;
 int a[MAXX];
 int cnt;

 int findpush(int n)
 {
     int flag = ;
     for(int i = ; i <= n; i++)
     {
         if(a[i] == )
         {
             flag = ;
             if(bel[i].size())
                 qu.push(bel[i]);
             else
                 cnt--;
             a[i] = INF;
         }
     }
     return flag;
 }

 int main()
 {
     int T, n, t, ed;
     scanf("%d", &T);
     while(T--)
     {
         scanf("%d",&n);
         cnt = n;

         for(int i = ; i <= n; i++)
             a[i] = , bel[i].clear();
         for(int i = ; i <= n; i++)
         {
             scanf("%d", &t);
            /* if(!t)
                 cnt--;*/
             while(t--)
             {
                 scanf("%d", &ed);
                 bel[ed].push_back(i);
                 a[i]++;
             }
         }
         while(qu.empty())
         {
             int flag = findpush(n);
             if(flag == )
             {
                 if(cnt)
                     printf("No\n");
                 else printf("Yes\n");
                 break;
             }
             int temp = qu.size();
             while(temp--)
             {

                 for(int i = ; i < (qu.front()).size(); i++)
                 {
                     a[(qu.front())[i] ]--;
                 }
                 qu.pop();
                 cnt--;
             }
         }
     }
 }