一道最小生成树的基本题吧,这里要注意两点
1.数组范围要开大,毕竟是生成树,开n平方即可
2.求边上权值算两点之间距离要注意精度的问题,多强制转换几次(double)防止WA
就OK了,剩下是Kruskal模板
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
int n,m,cnt;
double ans;
int x[1000005],y[1000005],fa[1000005];
struct Node{
int x;
int y;
double dis;
}a[1000005];
inline int Find(int i){
if(fa[i]==i)return i;
return fa[i]=Find(fa[i]);
}
inline void Union(int x,int y){
int f1=Find(x);
int f2=Find(y);
if(f1!=f2)fa[f1]=f2;
return;
}
inline bool cmp(Node a,Node b){
if(a.dis==b.dis)return a.x<b.x;
return a.dis<b.dis;
}
int main(){
scanf("%d%d",&n,&m);
for(register int i=1;i<=n;i++)
fa[i]=i;
for(register int i=1;i<=n;i++)
scanf("%d%d",&x[i],&y[i]);
for(register int i=1;i<=n;i++)
for(register int j=i+1;j<=n;j++){
a[++cnt].x=i;
a[cnt].y=j;
a[cnt].dis=(double)sqrt((double)(x[i]-x[j])*(double)(x[i]-x[j])+(double)(y[i]-y[j])*(double)(y[i]-y[j]));
}
int x,y;
for(register int i=1;i<=m;i++){
scanf("%d%d",&x,&y);
a[++cnt].x=x;
a[cnt].y=y;
a[cnt].dis=0.0;
}
sort(a+1,a+cnt+1,cmp);
for(register int i=1;i<=cnt;i++){
if(Find(a[i].x)!=Find(a[i].y)){
Union(a[i].x,a[i].y);
ans+=a[i].dis;
}
}
printf("%.2lf\n",ans);
return 0;
}
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原文:https://www.cnblogs.com/ainiyuling/p/11188388.html