Median Weight Bead（最短路—floyed傳遞閉包）

Description

There are N beads which of the same shape and size, but with different weights. N is an odd number and the beads are labeled as 1, 2, ..., N. Your task is to find the bead whose weight is median (the ((N+1)/2)th among all beads). The following comparison has been performed on some pairs of beads: 
A scale is given to compare the weights of beads. We can determine which one is heavier than the other between two beads. As the result, we now know that some beads are heavier than others. We are going to remove some beads which cannot have the medium weight. 

For example, the following results show which bead is heavier after M comparisons where M=4 and N=5. 

1.	Bead 2 is heavier than Bead 1.

2.	Bead 4 is heavier than Bead 3.

3.	Bead 5 is heavier than Bead 1.

4.	Bead 4 is heavier than Bead 2.

From the above results, though we cannot determine exactly which is the median bead, we know that Bead 1 and Bead 4 can never have the median weight: Beads 2, 4, 5 are heavier than Bead 1, and Beads 1, 2, 3 are lighter than Bead 4. Therefore, we can remove these two beads. 

Write a program to count the number of beads which cannot have the median weight. 

Input

The first line of the input file contains a single integer t (1 <= t <= 11), the number of test cases, followed by the input data for each test case. The input for each test case will be as follows: 
The first line of input data contains an integer N (1 <= N <= 99) denoting the number of beads, and M denoting the number of pairs of beads compared. In each of the next M lines, two numbers are given where the first bead is heavier than the second bead. 

Output

There should be one line per test case. Print the number of beads which can never have the medium weight.

Sample Input

1
5 4
2 1
4 3
5 1
4 2

Sample Output

2

題目意思：對於T組輸入輸出，有n個珍珠（n爲奇數），有m次稱重機會，排列在前面編號的珍珠比後面的珍珠要重，求出能判斷出有多少個珍珠的重量必定不
是中間值。

解題思路：在這裏咱們須要想明白的是要找的這個中間數，必定是重量要大於個數一半珍珠的重量或者是重量要小於個數一半珍珠的重量。
同時也要明白若是大於（小於）了一半的個數，必定不一樣時存在小於（大於）一半的個數。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int n,m;
int map[200][200];
int floyd()
{
    int i,j,k;
    for(k=1; k<=n; k++)
        for(i=1; i<=n; i++)
            for(j=1; j<=n; j++)
            {
                if(map[i][k]&&map[k][j])
                {
                    map[i][j]=1;///若是任意兩個點可以經過第三個點發生關係，那麼說明這兩個點也是有關係的
                }
            }
}
int main()
{
    int a,b,i,j,count,sum1,sum2,t,flag;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        memset(map,0,sizeof(map));
        flag=n/2;
        for(i=0; i<m; i++)
        {
            scanf("%d%d",&a,&b);
            map[a][b]=1;///能夠肯定關係的利用鄰接矩陣記錄爲1
        }
        floyd();
        count=0;
        for(i=1; i<=n; i++)
        {
            sum1=0;
            sum2=0;
            for(j=1; j<=n; j++)
            {
                if(map[i][j])///利用該矩陣求出比該點大的點
                {
                    sum1++;
                }
                if(map[j][i])///利用該矩陣求出比該點小的點
                {
                    sum2++;
                }

            }
            if(sum1>flag)///比該點大的點數超過了1/2，則這個點必定不是中值點
            {
                count++;
            }
            if(sum2>flag)///比該點小的點數超過了1/2，則這個點必定不是中值點
            {
                count++;
            }
        }
        printf("%d\n",count);
    }
    return 0;
}