Harry and Magical Computer
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1046 Accepted Submission(s): 424
Problem Description
In reward of being yearly outstanding magic student, Harry gets a magical computer. When the computer begins to deal with a process, it will work until the ending of the processes. One day the computer got n processes to deal with. We number the processes from 1 to n. However there are some dependencies between some processes. When there exists a dependencies (a, b), it means process b must be finished before process a. By knowing all the m dependencies, Harry wants to know if the computer can finish all the n processes.
Input
There are several test cases, you should process to the end of file.
For each test case, there are two numbers n m on the first line, indicates the number processes and the number of dependencies.1≤n≤100,1≤m≤10000
The next following m lines, each line contains two numbers a b, indicates a dependencies (a, b).1≤a,b≤n
For each test case, there are two numbers n m on the first line, indicates the number processes and the number of dependencies.
The next following m lines, each line contains two numbers a b, indicates a dependencies (a, b).
Output
Output one line for each test case.
If the computer can finish all the process print "YES" (Without quotes).
Else print "NO" (Without quotes).
If the computer can finish all the process print "YES" (Without quotes).
Else print "NO" (Without quotes).
Sample Input
3 23 12 13 33 22 11 3
Sample Output
YESNO
Source
BestCoder Round #25
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ac代码
#include<stdio.h>#include<string.h>int n,m;int dig[1010],map[1010][1010];int topu(){ int i,j; for(i=1;i<=n;i++) { int u=-1; for(j=1;j<=n;j++) { if(dig[j]==0) { dig[j]--; u=j; break; } } if(u==-1) return 0; for(j=1;j<=n;j++) { if(map[u][j]) { dig[j]--; map[u][j]=0; } } } return 1;}int main(){ while(scanf("%d%d",&n,&m)!=EOF) { int i; memset(map,0,sizeof(map)); memset(dig,0,sizeof(dig)); for(i=0;i<m;i++) { int u,v; scanf("%d%d",&u,&v); if(!map[u][v]) { map[u][v]=1; dig[v]++; } } if(topu()) { printf("YES\n"); } else printf("NO\n"); }}