題干：

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.

Input

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.

Output

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

Sample Input

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

Sample Output

YES NO

題目大意：

? ? ? 給出m對關系A-B，說明A是B的師傅，問輸入的關系是否合法。

解題報告：

? ? ?并查集是不可以的，因為不是同一種關系。判斷有向圖是否成環，考慮拓撲排序。

AC代碼：

#include<bits/stdc++.h>using namespace std; const int MAX = 500 + 5 ; int n,m; struct Node {int to;int w;int ne; } e[MAX]; int in[MAX],head[MAX],ans[MAX]; int cnt = 0,top = 0; priority_queue<int,vector<int > ,greater<int> > pq; bool topu() {//預處理 for(int i = 0; i<n; i++) {if(in[i] == 0) pq.push(i);}while(!pq.empty() ) {int cur = pq.top();pq.pop();ans[++top] = cur;for(int i = head[cur]; i!=-1; i=e[i].ne) {//是ne啊！！！！ in[e[i].to]--; if(in[e[i].to] == 0 ) pq.push(e[i].to); }}if(top != n) return false;else return true; } void init() {cnt = 0;top = 0;memset(in,0,sizeof(in));memset(head,-1,sizeof(head));while(!pq.empty() ) pq.pop(); } void add(int u,int v,int w) {e[cnt].to = v;e[cnt].w = w;e[cnt].ne = head[u];head[u] = cnt;cnt++; } int main() {int u,v;while(~scanf("%d%d",&n,&m) ) {if(n == 0) break;init();while(m--) {scanf("%d%d",&u,&v);add(u,v,0);in[v]++;}if(topu()) puts("YES");else puts("NO");}return 0 ; }

總結：

? ? ? i=e[i].ne這句別寫錯了就行，，別直接用他自動補全的i=e[i].to了。。。

? ??

總結

以上是生活随笔為你收集整理的【HDU - 3342】Legal or Not（拓扑排序）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。