題目1448：Legal or Not

題目1448：Legal or Not

時間限制：1 秒記憶體限制：128 兆特殊判題：否送出：3161解決：1480

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

問題分析：

代碼：

#include <iostream>
#include <stdio.h>
#include <vector>
#include <queue>
#define MAX 100
/* run this program using the console pauser or add your own getch, system("pause") or input loop */
using namespace std;
vector<int> edge[MAX];  //鄰接連結清單存儲圖 
queue<int> q;   //臨時存放入度為0的點 

int main(int argc, char** argv) {
    int inDegree[MAX] ; //統計每個結點的入度

    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF) {
        //1初始化
        for(int i=;i<n;i++) {
            //所有結點初始入度為0 
            inDegree[i] = ;
        }
        while(q.empty() == false){
            //清空隊列 
            q.pop();
        }

        //2.輸入邊  
        while(m--){
            //輸入m條邊
             int a,b;
             scanf("%d%d",&a,&b) ;
             inDegree[b]++; //b的入度加1         
             edge[a].push_back(b); 
        }

        //3.入度為0的結點加入到隊列中 
        for(int i=;i<n;i++){
            if(inDegree[i] == ){
                q.push(i);
            }
        }

        //4. 循環選擇入度為0的結點進行删除 直到沒有入度為0的點為止
        int ans = ; //記錄加入拓撲排序中的結點個數 
        while(q.empty() == false) {
            int tmp = q.front();
            q.pop();
            ans++; 
            //删除入度為0的點  即将與其相鄰的點組成的邊删除 入度減1
            for(int i=;i<edge[tmp].size();i++) {

                inDegree[edge[tmp][i]]--;
                if(inDegree[edge[tmp][i]] == ){
                    q.push(edge[tmp][i]);
                }
            }
        }

        if(ans == n){
            printf("YES\n");
        }else{
            printf("NO\n");
        }
    }
    return ;
}