HDU 1704  Rank （关系闭包）

Rank

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

Total Submission(s): 805 Accepted Submission(s):
279


[align=left]Problem Description[/align]
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).

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
For each test case, output a integer
which represent the max possible number of queries that you can't
tell lcy.

 

[align=left]Sample Input[/align]

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

 

[align=left]Sample Output[/align]

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.

 

[align=left]Source[/align]

2007省赛集训队练习赛（1）

 

[align=left]Recommend[/align]
lcy
 
关闭闭包 floyd算法
 
代码：

#include<stdio.h>

#include<string.h>
#define MAXN 501

int map[MAXN][MAXN],n;
void floyd()

{

    int
i,j,k;

   
for(k=1;k<=n;k++)

       
for(i=1;i<=n;i++)

           
if(map[i][k])

           
{

               
for(j=1;j<=n;j++)

                   
map[i][j]=map[i][j]||map[i][k]&&map[k][j];

           
}

}

int main()

{

    int
i,j,T,m,a,b,ans;

   
scanf("%d",&T);

   
while(T--)

    {

       
memset(map,0,sizeof(map));

       
scanf("%d%d",&n,&m);

       
while(m--)

       
{

           
scanf("%d%d",&a,&b);

           
map[a][b]=1;

       
}

       
floyd();

       
ans=0;

       
for(i=1;i<=n;i++)

           
for(j=i+1;j<=n;j++)

               
if(!(map[i][j]||map[j][i]))

                   
ans++;

               
printf("%d\n",ans);

               

    }

    return
0;

}