HDU 1704 Rank【传递闭包】

解题思路：给出n个选手，m场比赛，问不能判断胜负的询问最多有多少种

用传递闭包即可 但是如果直接用3重循环会超时 在判断d[i][j]=d[i][k]||d[k][j]是否连通的时候 可以加一个if语句判断一下d[i][k]是否为1，为1再进行第三重循环，不为1则不进行第三次循环

 反思：例如询问 3和1,1和3是相同的情况，所以最后判断的时候，只需要判断上三角矩阵即可。

Rank

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1134    Accepted Submission(s): 429

Problem Description

there are N ACMers in HDU team. ZJPCPC Sunny Cup 2007 is coming, and lcy want to select some excellent ACMers to attend the contest. There have been M matches since the last few days(No two ACMers will meet each other at two matches, means between two ACMers there will be at most one match). lcy also asks"Who is the winner between A and B?" But sometimes you can't answer lcy's query, for example, there are 3 people, named A, B, C.and 1 match was held between A and B, in the match A is the winner, then if lcy asks "Who is the winner between A and B", of course you can answer "A", but if lcy ask "Who is the winner between A and C", you can't tell him the answer. As lcy's assistant, you want to know how many queries at most you can't tell lcy(ask A B, and ask B A is the same; and lcy won't ask the same question twice).

 

Input

The input contains multiple test cases. The first line has one integer,represent the number of test cases. Each case first contains two integers N and M(N , M <= 500), N is the number of ACMers in HDU team, and M is the number of matchs have been held.The following M lines, each line means a match and it contains two integers A and B, means A wins the match between A and B.And we define that if A wins B, and B wins C, then A wins C.

 

Output

For each test case, output a integer which represent the max possible number of queries that you can't tell lcy.

 

Sample Input

3 3 3 1 2 1 3 2 3 3 2 1 2 2 3 4 2 1 2 3 4

 

Sample Output

0 0 4

Hint

in the case3, if lcy ask (1 3 or 3 1) (1 4 or 4 1) (2 3 or 3 2) (2 4 or 4 2), then you can't tell him who is the winner.

 

#include<stdio.h>
#include<string.h>
int d[505][505],a[505][505];
int main()
{
	int ncase,n,m,i,j,k,sum,x,y;
	scanf("%d",&ncase);
	while(ncase--)
	{
		memset(d,0,sizeof(d));
		sum=0;
		scanf("%d %d",&n,&m);
		while(m--)
		{
			scanf("%d %d",&x,&y);
			d[x][y]=1;
		}
		
		for(k=1;k<=n;k++)	
		for(i=1;i<=n;i++)
		{
			if(d[i][k])
	        for(j=1;j<=n;j++)			
			 d[i][j]=d[i][j]||(d[k][j]);		    
		}
			
		for(i=1;i<=n;i++)
			for(j=i+1;j<=n;j++)
				if(!(d[i][j]||d[j][i]))
				sum++;
		printf("%d
",sum);
	}	
}