poj 1975 Median Weight Bead(传递闭包 Floyd)
题意:n个珠子,给定它们之间的重量关系。按重量排序。求确定肯定不排在中间的珠子的个数
分析:由于n为奇数。中间为(n+1)/2,对于某个珠子。若有至少有(n+1)/2个珠子比它重或轻,则它肯定不排在中间
能够将能不能确定的权值初始化为0,能确定重量关系的权值设为1
#include<stdio.h>
#include<string.h>
int a[110][110];
int main()
{
int T,n,m,i,j,k,d,x,sum;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
memset(a,0,sizeof(a));
while(m--){
scanf("%d%d",&i,&j);
a[i][j]=1;
}
for(k=1;k<=n;k++)
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
if(a[i][k]&&a[k][j]) //当i比k重,k比j重,则i比j重
a[i][j]=1;
sum=0;
for(i=1;i<=n;i++){
d=x=0;
for(j=1;j<=n;j++){
if(a[i][j])
d++;
else if(a[j][i])
x++;
}
if(d>=(n+1)/2||x>=(n+1)/2)
sum++;
}
printf("%d\n",sum);
}
return 0;
}
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