Call to your teacher

Call to your teacher

x有y的电话号码，也就是说x可以给y打，y可以给所拥有的号码打，依此类推，最后有没有一个人有n的号码

单向的结构，代码呢？

#include <bits/stdc++.h>
using namespace std;
int fa[60],n,m;
int main(){
    ios::sync_with_stdio(0);
    while(cin >> n >> m){
        fa[1] = 1;
        for(int i = 2; i <= n; i++)
            fa[i] = 0;
        for(int i = 0; i < m; i++){
            int x,y;
            cin >> x >> y;
            if(fa[x]) fa[y] = 1;
        }
        if(fa[n]) cout << "Yes" << endl;
        else cout << "No" << endl;
    }
    return 0;
}

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