HDU 3342 legal or not (topological sorting)

Legal or not

Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 3151 accepted submission (s): 1432


Problem descriptionacm-DIY is a large QQ group where your excellent acmers get together. it is so harmonious that just like a big family. every day, daily "holy cows" like HH, HH, AC, ZT, KP, BF, qinz and so on chat on-line to exchange their ideas. when someone has questions,
Export 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 using pairs of "master and Prentice ". but then problem occurs: there are too processing masters and too
When prentices, how can we know whether it is legal or not?

We all know a master can have tables 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 3 Xian for instant, HH is 3xian's master and, at the same time, 3 Xian
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.

 

Inputthe 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.

Outputfor each test case, print in one line the judgement of the messy relationship.
If it is legal, output "yes", otherwise "no ".

Sample Input

3 20 11 22 20 11 00 0

 

Sample output

YESNO

Question: determine the relationship between the master and the apprentice, and determine whether the relationship is correct or not.

Solution: in fact, it is to determine whether or not the topology can be sorted, investigate the nature of the topological sorting, and use DFS to collect and judge the data.

#include<stdio.h>#include<string.h>#define N 105int n, m;int vis[N], map[N][N];int DFS(int k){vis[k] = 1;for (int i = 0; i < n; i++){if (map[k][i]){if(vis[i])return 1;else if(DFS(i))return 1;}}vis[k] = 0;return 0;}int main(){while (scanf("%d%d", &n, &m), n || m){// Init.memset(map, 0, sizeof(map));int ok = 0;// Read.for (int i = 0; i < m; i++){int a, b;scanf("%d%d", &a, &b);map[b][a] = 1;}// DFS.for (int i = 0; i < n; i++){memset(vis, 0, sizeof(vis));if (DFS(i)){ok = 1;break;}}if (!ok)printf("YES\n");elseprintf("NO\n");}return 0;}