题目1448：Legal or Not 【拓扑排序】

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 20 11 22 20 11 00 0

样例输出：

YESNO

代码：

#include <stdio.h>#include <vector>#include <queue>using namespace std;vector<int> vi[101];queue<int> Q;int degree[101];int main() {    int n,m;    while(scanf("%d %d",&n,&m)!=EOF) {if(n == 0) break;for(int i=0;i<n;i++) {    vi[i].clear();    degree[i] = 0;}while(m--) {    int x,y;    scanf("%d %d",&x,&y);    degree[y]++;    vi[x].push_back(y);}while(!Q.empty()) Q.pop();for(int i=0;i<n;i++) {    if(degree[i] == 0)Q.push(i);}int cnt = 0;while(!Q.empty()) {    int now = Q.front();    Q.pop();    cnt++;    for(int i=0;i<vi[now].size();i++) {degree[vi[now][i]]--;if(degree[vi[now][i]] == 0)    Q.push(vi[now][i]);    }}if(cnt == n)    puts("YES");else    puts("NO");    }    return 0;}

对于一个有向无环图（DAG图），对于所有的有向边(U,V)(由U指向V)，在该序列中结点U都排列在结点V之前。

满足该要求的结点序列，被称为满足拓扑次序的序列。求这个序列的过程就是拓扑排序。

先选择 一个入度为0的结点，然后删去，一直重复，知道所有结点都删去。若找不到入度为0的结点了，则说明有环路。

无论何时，当需要判断某个图是否是有向无环图时，我们都需要立刻联想到拓扑排序！

标准模板std::queue。

1.queue<int> Q;

2.Q.push(x);

3.x = Q.front();

4.Q.pop();

5.Q.empty();