ZZULI 1599（POJ 1975）Median Wei…

Median Weight Bead

Time Limit: 1000MSMemory Limit: 30000KTotal Submissions: 2182Accepted: 1076

Description

There are N beads which of the same shape andsize, but with different weights. N is an odd number and the beadsare labeled as 1, 2, ..., N. Your task is to find the bead whoseweight is median (the ((N+1)/2)th among all beads). The followingcomparison has been performed on some pairs of beads:
A scale is given to compare the weights of beads. We can determinewhich one is heavier than the other between two beads. As theresult, we now know that some beads are heavier than others. We aregoing to remove some beads which cannot have the mediumweight.

For example, the following results show which bead is heavier afterM 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 isthe median bead, we know that Bead 1 and Bead 4 can never have themedian 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 twobeads.

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

Input

The first line of the input file contains asingle integer t (1 <= t <= 11), thenumber of test cases, followed by the input data for each testcase. 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 ofbeads, and M denoting the number of pairs of beads compared. Ineach of the next M lines, two numbers are given where the firstbead is heavier than the second bead.

Output

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

Sample Input

15 42 14 35 14 2

Sample Output

2

Source

Tehran Sharif 2004 Preliminary

 

这题虽然是Floyd，可是我怎么看怎么都有种暴力的感觉，16MS水过……

 

题意：给一组数据，表示前面的大于后面的，然后如果从小到大排列的话第（n+1）/2个就是中间数，现在问根据数据有几个不可能是中间数

 

这题猛一看貌似挺简单的，可是真去实现的时候却又无从下手了（好纠结啊），赛后看了人家的才知道是Floyd算法，于是搜索各种解题报告啊，然后在各种纠结下看懂了怎么做的了，不会的参考Floyd算法……

 

代码：

C语言: 高亮代码由发芽网提供

#include<stdio.h>
#include<string.h>
intmain()
{
    int light[100][100],weight[100][100],i,j,k,m,n,a,b,Case;
    int ans1,ans2,num;
    scanf("%d",&Case);
    while(Case--){
       scanf("%d%d",&n,&m);
       memset(light,0,sizeof(light));
       memset(weight,0,sizeof(weight));
       for(i=0;i<m;i++){
           scanf("%d%d",&a,&b);
           light[a][b]=1;
           weight[b][a]=1;
       }

       for(k=1;k<=n;k++)
           for(i=1;i<=n;i++)
               for(j=1;j<=n;j++)
               {
                   if(light[i][k]&&light[k][j])
                       light[i][j]=1;
                   if(weight[i][k]&&weight[k][j])
                       weight[i][j]=1;
               }
       num=0;
       for(i=1;i<=n;i++)
       {
           ans1=0;ans2=0;
           for(j=1;j<=n;j++)
           {
               if(light[i][j])
                   ans1++;
               if(weight[i][j])
                   ans2++;
           }
           if(ans1>=(n+1)/2)
               num++;
           if(ans2>=(n+1)/2)
               num++;
       }
       printf("%d\n",num);
    }
    return 0;
}

PS：原来的三重循环是按照先i后j然后k循环的，但是提交WA，至今还没有理解，哪位大牛知道了麻烦给留个言