POJ 3625 最小生成树 Prim C++

Building Roads

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11861		Accepted: 3376

Description

Farmer John had just acquired several new farms! He wants to connect the farms with roads so that he can travel from any farm to any other farm via a sequence of roads; roads already connect some of the farms.

Each of the N (1 ≤ N ≤ 1,000) farms (conveniently numbered 1..N) is represented by a position (X_i, Y_i) on the plane (0 ≤ X_i ≤ 1,000,000; 0 ≤ Y_i ≤ 1,000,000). Given the preexisting M roads (1 ≤ M ≤ 1,000) as pairs of connected farms, help Farmer John determine the smallest length of additional roads he must build to connect all his farms.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Two space-separated integers: X_i and Y_i 
* Lines N+2..N+M+2: Two space-separated integers: i and j, indicating that there is already a road connecting the farm i and farm j.

Output

* Line 1: Smallest length of additional roads required to connect all farms, printed without rounding to two decimal places. Be sure to calculate distances as 64-bit floating point numbers.

Sample Input

4 1
1 1
3 1
2 3
4 3
1 4

Sample Output

4.00

一个最小生成树问题，kruskal算法会TLE

没什么可说i的，这玩意得存模板，这题唯一的不一样只是改变了权值为两点间距。

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<queue>
#define INF 0x3f3f3f3f
using namespace std;
int n,m;
 struct node
 {
     double x,y;
 }a[1005];
 int vis[1005];
 double d[1005][1005];
double dis(node a,node b)
 {
     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
 }
 double prim()
 {
     memset(vis,0,sizeof(vis));
     double low[1005];
     int pos=1;
     double ans=0;
     vis[1]=1;
     for(int i=2;i<=n;i++){
        low[i]=d[pos][i];

     }
     for(int i=1;i<n;i++){
        double min=INF;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j]&&min>low[j]){
                min=low[j];
                pos=j;
            }
        }
        vis[pos]=1;
        ans+=min;
        for(int i =1;i<=n;i++)
        {
            if(!vis[i]&&low[i]>d[pos][i]){
                low[i]=d[pos][i];
            }
        }
     }
     return ans;
 }
 int main()
 {
     while(~scanf("%d%d",&n,&m))
     {
         memset(d,INF,sizeof(d));
         for(int i=1;i<=n;i++)
         {
             scanf("%lf%lf",&a[i].x,&a[i].y);
         }
         for(int i=1;i<=n;i++){
            for(int j =i+1;j<=n;j++){
                d[i][j]=d[j][i]=dis(a[i],a[j]);
            }
         }
         for(int i=0;i<m;i++){
                int x,y;
            scanf("%d%d",&x,&y);
         d[x][y]=0;
         d[y][x]=0;
         }
         printf("%.2f\n",prim());
     }
 }