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(100010);
queue< vector<int> >qu;
int a[MAXX];
int cnt;

int findpush(int n)
{
    int flag = 1;
    for(int i = 1; i <= n; i++)
    {
        if(a[i] == 0)
        {
            flag = 0;
            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 = 1; i <= n; i++)
            a[i] = 0, bel[i].clear();
        for(int i = 1; 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 == 1)
            {
                if(cnt)
                    printf("No\n");
                else printf("Yes\n");
                break;
            }
            int temp = qu.size();
            while(temp--)
            {

                for(int i = 0; i < (qu.front()).size(); i++)
                {
                    a[(qu.front())[i] ]--;
                }
                qu.pop();
                cnt--; 
            }        
        }
    }
}