hdu 1704 Rank (传递闭包)

Rank

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

Total Submission(s): 990    Accepted Submission(s): 361

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.

   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    ps:此题貌似是传递闭包 反正我用搜索和floyd()做不了 （用floyd()会超时） 不过其实传递闭包就是借用的floyd()思想
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    
    
    
    
    
#include<stdio.h>
#include<math.h>
#include<string.h>
#define maxn 505

int city[maxn][maxn];
int n,m,ans;

void floyd()    // 改良版 变为传递闭包
{
    int i,j,k;
    for(i=1; i<=n; i++)
    {
        for(j=1; j<=n; j++)
        {
            if(city[j][i])     // 如果j,i 连通  如果这里变为city[i][j] 你会发现会WA
            // 因为会会漏掉很多win关系, 主要是历遍的顺序不一样造成的 不信的话不妨试一下我贴的数据
            {
                for(k=1; k<=n; k++)
                {
                    if(city[i][k]) city[j][k]=1;   // 如果i,k连通 则j,k连通
                }
            }
        }
    }
}
void solve()
{
    int i,j;
    ans=0;
    for(i=1; i<=n; i++)
    {
        for(j=i+1; j<=n; j++)
        {
            if(!city[i][j]&&!city[j][i]) ans++;
        }
    }
}
int main()
{
    int i,j,t,l,r;
    scanf("%d",&t);
    while(t--)
    {
        memset(city,0,sizeof(city));
        scanf("%d%d",&n,&m);
        for(i=1; i<=m; i++)
        {
            scanf("%d%d",&l,&r);
            city[r][l]=1;
        }
        floyd();
        solve();
        printf("%d\n",ans);
    }
    return 0;
}
/*
这组数据可以测试第16行为什么是city[j][i]
1
8 8
1 2
2 3
3 4
4 5
6 5
7 5
8 7
4 8
*/