标签:
并查集+最小生成树
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7669 Accepted Submission(s):
3882
#include<stdio.h>
#include<algorithm>
#include<math.h>
using namespace std;
int set[110];
struct record//注意此题要用实型
{
double a;
double b;
double ju;//两点之间距离
}s[10000];
int find(int fa)
{
int ch=fa;
int t;
while(fa!=set[fa])
fa=set[fa];
while(ch!=fa)
{
t=set[fa];
set[ch]=fa;
ch=t;
}
return fa;
}
void mix(int x,int y)
{
int fx,fy;
fx=find(x);
fy=find(y);
if(fx!=fy)
set[fx]=fy;
}
double dis(double x1,double y1,double x2,double y2)
{
return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)); //计算两点之间距离
}
bool cmp(record c,record d)
{
return c.ju<d.ju;
}
int main()
{
int m,j,i,point;
double sum;
double a[110];//这两个数组用来储存点的坐标,a数组储存横标
double b[110];//b数组储存纵标
while(scanf("%d",&point)!=EOF)
{
for(i=0;i<=point;i++)
set[i]=i;
for(i=1;i<=point;i++)
{
scanf("%lf%lf",&a[i],&b[i]);
}
m=0;
for(i=1;i<=point-1;i++) //此循环求任意两点之间距离
{ //并记录下此两点位置
for(j=i+1;j<=point;j++)
{
s[m].ju=dis(a[i],b[i],a[j],b[j]);
s[m].a=i;
s[m].b=j;
m++;
}
}
sort(s,s+m,cmp);
sum=0;
for(i=0;i<m;i++)
{
if(find(s[i].a)!=find(s[i].b))
{
mix(s[i].a,s[i].b);
sum+=s[i].ju;
}
}
printf("%.2lf\n",sum);
}
return 0;
}
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原文地址:http://www.cnblogs.com/tonghao/p/4470938.html