Legal or Not(拓扑排序)
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Legal or Not
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9200 Accepted Submission(s): 4275
Problem Description
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.
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.
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".
If it is legal, output "YES", otherwise "NO".
Sample Input
3 20 11 22 20 11 00 0
Sample Output
YESNO题目大意:a是b的师傅,通过拓扑排序的步骤,如果能正确操作完所有步骤就输出YES(一会详解,这里说的比较浅),否则输出NO。
这里a是b的师傅,b不能是a的师傅,如果b是a的师傅要输出NO(题目样例2可看出)这里b是a的师傅表示与全序关系概念相反(矛盾),
就是说题意就是要找是否有这样的矛盾,如果存在就输出NO。
#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>using namespace std;int dree[1001];int map[1001][1001];int n,m;int main(){ int i,j; while(scanf("%d %d",&n,&m)!=EOF,n,m) { memset(map,0,sizeof(map)); memset(dree,0,sizeof(dree)); for(i=0;i<m;i++) { int x,y; scanf("%d %d",&x,&y); if(map[x][y]==0) { map[x][y]=1; dree[y]++; } } int cont=0; int flag=0; while(1) { int v; int ok=1; for(i=0;i<n;i++) { if(dree[i]==0) { v=i; ok=0; break; } } cont++; dree[v]=-1; if(ok==1) { flag=1; break; } if(cont==n) { break; } for(i=0;i<n;i++) { if(map[v][i]==1) { map[v][i]=-1; dree[i]--; } } } if(flag==1) { printf("NO\n"); } else { printf("YES\n"); } } return 0;}
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