對一個有向無環圖(Directed Acyclic Graph簡稱DAG)G進行拓撲排序，是將G中所有頂點排成一個線性序列，使得圖中任意一對頂點u和v，若邊(u,v)∈E(G)，則u在線性序列中出現在v之前。

通常，這樣的線性序列稱為滿足拓撲次序(Topological Order)的序列，簡稱拓撲序列。簡單的說，由某個集合上的一個偏序得到該集合上的一個全序，這個操作稱之為拓撲排序。

執行步驟

例5.7 Leagal or Not (1448)

題目描述：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.

輸入：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.

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

樣例輸入：
3 2
0 1
1 2
2 2
0 1
1 0
0 0
樣例輸出：
YES
NO

#include<iostream>
#include<vector>
#include<queue>
#include<stdio.h>
using namespace std;
vector<int> edge[501];
queue<int>Q;

int main(){
    int inDegree[501];
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        if(n==0&&m==0) break;
        for(int i=0;i<n;i++){
            inDegree[i]=0;
            edge[i].clear();
        }
        while(!Q.empty()) Q.pop();
        while(m--){
            int a,b;
            scanf("%d%d",&a,&b);
            inDegree[b]++;
            edge[a].push_back(b);
        }
        for(int i=0;i<n;i++)
            if(inDegree[i]==0)
                Q.push(i);
        int cnt=0;
        while(!Q.empty()){
            int nowP=Q.front();
            Q.pop();
            cnt++;
            for(int i=0;i<edge[nowP].size();i++){
                inDegree[edge[nowP][i]]--;
                if(inDegree[edge[nowP][i]]==0)
                    Q.push(edge[nowP][i]);
            }
        }
        if(cnt==n) puts("YES");
        else puts("NO");
    }
    return 0;
}

該代碼所有節點至多進入隊列一次，但在每個結點被取出時我們都要遍歷以其為弧尾的邊，故復雜度為O（N+E），其中N為結點的個數，E為邊的個數。

圖論_拓撲排序