HDU 5154_Harry and Magical Computer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1676    Accepted Submission(s): 656

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

 

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).

 

Sample Input

3 23 12 13 33 22 11 3

 

Sample Output

YESNO

这道题的意思是判断给出的数值里面有没有环，有的话输出NO，没有输出YES。

其实一开始是想到用二维数组来弄的，后来果断爆内存了，再后来想到了用一维数组来代替二维数组，一开始初始化数组的时候让数组的值等于它的下标，然后再输入数值，不断更新一维数组，如果遇到一开始更新时的那个值，就说明输入的数组里面有环，输出NO，否则输出YES。

#include<iostream>#include<string.h>#include<stdio.h>using namespace std;int num[10010];int flag=0;int find(int a,int b){if(num[b]==a)  return 1;if(num[b]==b)  return 0;find(a,num[b]);}int main(){int n,m;while(scanf("%d %d",&n,&m)!=EOF){flag=0;for(int i=1;i<=n;i++)  num[i]=i;int ans,nut;for(int i=1;i<=m;i++){scanf("%d %d",&ans,&nut);if(find(ans,nut))  flag=1;else  num[ans]=nut;}if(flag==1)  printf("NO\n");else  printf("YES\n");}return 0;}