HDU 5154 Harry and Magical Computer（找环）

Harry and Magical Computer

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1743 Accepted Submission(s): 684



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 2
3 1
2 1
3 3
3 2
2 1
1 3

Sample Output

YES
NO

Source

BestCoder Round #25

问题描述

作为年度优秀魔法学员的奖赏，哈利得到了一台具有魔力的计算机。这台计算机一旦开始处理某个任务，就会一直处理到这个任务结束为止(所以你可以认为它是单线程的)。有一天，这台计算机得到了n个任务要处理，分别标号1到n。这n个任务之间又有一些依赖关系，假如存在依赖关系(a, b)，那么要处理a任务，必须先将b任务完成。现在哈利得到了所有的这些依赖关系，一共m个。他想知道，这台计算机能否完成所有的任务。

输入描述

多组输入数据
每组数据第一行由n m组成，分别代表任务数和依赖关系数。1 \leq n \leq 100, 1 \leq m \leq 100001≤n≤100,1≤m≤10000
接下来m行，每行两个数a b，表示一组依赖关系。1 \leq a, b \leq n1≤a,b≤n

输出描述

每组输出一行，"YES"(没有引号)表示能完成所有任务。
"NO"(没有引号)表示不能。

输入样例

3 2
3 1
2 1
3 3
3 2
2 1
1 3

输出样例

YES
NO

解题思路：如果存在任务不能完成，那么说明该任务既出现在某任务的前面又出现在某任务的后面，换句话说就是某些任务与该任务一起组成了一个环。用弗洛伊德算法，判断最后的邻接矩阵的对角线是否有1存在，如果有，说明存在环，否则没有。

代码如下：

#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <string>
#include <algorithm>
#include <vector>
#include <deque>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <sstream>
#include <fstream>
#include <limits.h>
#include <ctime>
#define debug "output for debug\n"
#define pi (acos(-1.0))
#define eps (1e-6)
#define inf (1<<28)
#define sqr(x) (x) * (x)
#define mod 1000000007
using namespace std;
typedef long long ll;
typedef unsigned long long ULL;
int a[105][105];
int main()
{
int i,j,k,n,m,x,y,t;
while(~scanf("%d%d",&n,&m))
{
memset(a,0,sizeof(a));
for(i=1;i<=m;i++)
{
scanf("%d%d",&x,&y);
a[y][x]=1;
}
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
for(k=1;k<=n;k++)
{
if(a[i][j]<min(a[i][k],a[k][j]))
a[i][j]=min(a[i][k],a[k][j]);
}
}
}
int flag=0;
for(i=1;i<=n;i++)
{
if(a[i][i]==1)
{
flag=1;
break;
}
}
if(flag)
printf("NO\n");
else
printf("YES\n");
}
return 0;
}