nyoj 170 网络的可靠性

时间:2023-05-04 11:50:02

题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=170

思路:统计每个节点的度,将度为1的节点消去所需要的最少的边即为答案。



代码:

#include "stdio.h"  //nyoj 170 网络的可靠性
#include "string.h"
#define N 10005
int du[N]; int main()
{
int n;
int i;
int x,y;
while(scanf("%d",&n)!=EOF)
{
if(n==1 || n==2) {printf("0\n"); continue; }
memset(du,0,sizeof(du));
for(i=1; i<n; ++i)
{
scanf("%d %d",&x,&y);
du[x]++;
du[y]++;
}
int num=0;
for(i=1; i<=n; ++i)
{
if(du[i]==1)
num++;
}
if(num%2==0)
num = num/2;
else
num = num/2+1;
printf("%d\n",num);
}
return 0;
}