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double 在HDU可以过啊 精度也考虑了 修改成int 完全正确 过了 看来double得少用啊#include<stdlib.h>
#include<stdio.h>
#include<string.h>
#define eps 1e-10
#define maxn 110 //总最多顶点个数
#define inf 9999999 //无穷大
double map[maxn][maxn],disk[maxn];//顶点之间的关系-map邻接矩阵
double prim(int n,int s)
{
int visit[maxn],i,w,j;
double min=inf,sum=0;
memset(visit,0,sizeof(visit));
visit[s]=1;
for(i=0;i<n;i++)
disk[i]=map[0][i];
for(i=1;i<n;i++)
{
min=inf;w=-1;
for( j=1;j<n;j++)
{
if(!visit[j]&&(min-disk[j]>=eps))
{min=disk[j];w=j;}
}
if(w==-1)return -1;
visit[w]=1;
sum+=min;
for(j=1;j<n;j++)
{
if(!visit[j]&&(disk[j]-map[w][j]>=eps))
disk[j]=map[w][j];
}
}
return sum;
}
double prim_solve(int n,int s)//使用prim 函数
{
double sum=0;
sum=prim(n,s);
return sum;
}
int main()
{
int n,i,j,q;
int a,b;
double sum,x;
while(scanf("%d",&n)!=EOF)
{sum=0;
//init();
for(i=0;i<n;i++)
for(j=0;j<n;j++)
{
scanf("%lf",&x);
map[i][j]=x;
}
scanf("%d",&q);
for(i=0;i<q;i++)
{
scanf("%d%d",&a,&b);
map[a-1][b-1]=map[b-1][a-1]=0;
}
sum = prim( n, 0);
printf("%.0lf\n",sum);
}
return 0;
}
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