HUST——1103Party（拓扑排序+个人见解）

1103: Party

Time Limit: 2 Sec  Memory Limit: 64 MB
Submit: 11  Solved: 7

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. 

今天找了篇博客认真看了下拓扑排序，想自己重新写下这道题。一次就过了运气不错。（很多东西重新回过头来看会有更深入的理解和发现）仔细思考一下确实如文章所说：

不难看出该算法的实现十分直观，关键在于需要维护一个入度为0的顶点的集合：

每次从该集合中取出(没有特殊的取出规则，随机取出也行，使用队列/栈也行，下同)一个顶点，将该顶点放入保存结果的List中。

紧接着循环遍历由该顶点引出的所有边，从图中移除这条边，同时获取该边的另外一个顶点，如果该顶点的入度在减去本条边之后为0，那么也将这个顶点放到入度为0的集合中。然后继续从集合中取出一个顶点…………

当集合为空之后，检查图中是否还存在任何边，如果存在的话，说明图中至少存在一条环路。不存在的话则返回结果List，此List中的顺序就是对图进行拓扑排序的结果。

首先要知道一点入度和出度的概念，比如我在学数学之前要学语文，此时语文和数学之间有一个箭头，那按正常思维肯定是语文比数学优先吧，因为你肯定要先学语文才能学数学啊，那么就在语文和数学之间连一条有向边，就像这样语文——>数学，可以当成你玩游戏时的进度，过了语文这个任务才能过数学。嗯这样应该非常好理解了吧。

因此需要用到三个东西使得程序易于实现和理解

1、容器数组：是记录每一个顶点a所指向的顶点bi的集合；

比如1——>2那么我们就在vector[1]中压入2这个点。

2、入度数组：记录每一个顶点的入度。显然例子中rudu[数学]=1;

3、维护每次找到的入度为0的队列：拓扑排序可以看成一种模拟——首先找到入度为0的即没人可以制约它的点，然后将这个点和它指出去的边删去，此时它连到的点入度肯定会因此减1（平凡图情况下）。那么很可能此时又会产生一个/多个入度为0的点。然后又可以重复前面步骤了。直到找不到入度为0的点或者所有边已经访问过。

那拓扑归拓扑，排序是什么鬼？就是每次出现的入度变为0的点按照出现的先后顺序构成的一个序列。用容器或者数组加一个尾指针来保存就可以。当然此题不需要输出排序结果。

代码（注释内容可以输出排序结果）：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#define INF 0x3f3f3f3f
#define MM(x) memset(x,0,sizeof(x))
using namespace std;
typedef long long LL;
int const N=100010;
int rudu[N];
vector<int>V[N];
inline void init(int n)
{
	for (int i=0; i<=n; i++)
	{
		V[i].clear();
	}
	MM(rudu);
}
int main(void)
{
	int tcase,i,j,a,n,m,sum;
	scanf("%d",&tcase);
	while (tcase--)
	{
		queue<int>Q;
		scanf("%d",&n);
		sum=0;init(n);
		for (i=1; i<=n; i++)
		{
			scanf("%d",&m);
			if(m==0)
			{
				Q.push(i);
				continue;
			}	
			for (j=1; j<=m; j++)
			{
				scanf("%d",&a);
				V[a].push_back(i);
			}
			rudu[i]+=m;
			sum+=m;
		}
		//vector<int>ans;
		while (!Q.empty())
		{
			int now=Q.front();
			Q.pop();
			int sz=V[now].size();
			for (i=0; i<sz; i++)
			{
				int v=V[now][i];
				rudu[v]--;
				sum--;
				if(rudu[v]==0)
				{
					Q.push(v);
					//ans.push_back(v);
				}
			}
		}
		if(sum)
			puts("No");
		else
		{
			puts("Yes");
			/*for (vector<int>::iterator it=ans.begin(); it!=ans.end(); it++)
			{
				printf("%d ",*it);
			}
			putchar('
');*/
		}	
	}
	return 0;
}