九度OJ 题目1448：Legal or Not

题目描述：

ACM-DIY is a large QQ group where many excellent acmers get together. It is so harmonious that just like a big family. Every day,many “holy cows” like HH, hh, AC, ZT, lcc, BF, Qinz and so on chat on-line to exchange their ideas. When someone has questions, many warm-hearted cows like Lost will come to help. Then the one being helped will call Lost “master”, and Lost will have a nice “prentice”. By and by, there are many pairs of “master and prentice”. But then problem occurs: there are too many masters and too many prentices, how can we know whether it is legal or not?We all know a master can have many prentices and a prentice may have a lot of masters too, it’s legal. Nevertheless，some cows are not so honest, they hold illegal relationship. Take HH and 3xian for instant, HH is 3xian’s master and, at the same time, 3xian is HH’s master,which is quite illegal! To avoid this,please help us to judge whether their relationship is legal or not. Please note that the “master and prentice” relation is transitive. It means that if A is B’s master ans B is C’s master, then A is C’s master.

输入：

The input consists of several test cases. For each case, the first line contains two integers, N (members to be tested) and M (relationships to be tested)(2 <= N, M <= 100). Then M lines follow, each contains a pair of (x, y) which means x is y’s master and y is x’s prentice. The input is terminated by N = 0.TO MAKE IT SIMPLE, we give every one a number (0, 1, 2,…, N-1). We use their numbers instead of their names.

输出：

For each test case, print in one line the judgement of the messy relationship.If it is legal, output “YES”, otherwise “NO”.

样例输入：

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

样例输出：

YES
NO

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不合法情况的发生原因：有向图中有圈
怎么确定有向图中有没有圈：拓扑排序，有圈的排不出的~
怎么进行拓扑排序：
找入读为0的节点n，删掉节点及节点的关联边，然后n节点放到拓扑排序的队列中
具体还是看代码
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#include
using namespace std;
int findZeroInDegree(int **adjMatrix, int N, bool *visited)
{
//在邻接矩阵中找【入】度为0的节点。
for(int col=0; col> N && N!=0)
{
int M;
cin >> M;
int **adjMatrix = new int*[N];
bool *visited = new bool[N];
int i;
for(i=0; i> x >> y;
adjMatrix[x][y] = 1;
}
///
int count = 0;
int nextNode = findZeroInDegree(adjMatrix, N, visited);
while(nextNode != -1)
{
removeRelatedEdge(adjMatrix, nextNode, N);
nextNode = findZeroInDegree(adjMatrix, N, visited);
count ++;
}
///
if(count == N)
cout << "YES" << endl; else cout << "NO" << endl; ///Delete for(i=0; i