tzoj4679 Building Roads（最小生成树+Prim算法）

时间限制(普通/Java):1000MS/3000MS     内存限制:65536KByte
总提交: 245            测试通过:79

描述

 

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.

输入

 

* 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.

输出

 

* 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.

样例输入

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

样例输出

 4.00

题目来源

USACO 2007 December Silver

 

 

tzoj卡掉了Kruskal算法，只能交Prim

#include <bits/stdc++.h>
using namespace std;

int n,m;
const int MAX=1005;
const double INF=0x3f3f3f3f;
struct Point{
    double x,y;
}p[MAX];
double g[MAX][MAX];
bool vis[MAX];
double mind[MAX];

inline double DST(Point a,Point b){
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

double prim(){
    double res=0,minn;
    int k;
    vis[1]=true;
    for(int i=1;i<=n;i++){
        mind[i]=g[1][i];
    }
    for(int i=1;i<=n-1;i++){
         minn=INF;
         for(int j=1;j<=n;j++){
            if(!vis[j]&&mind[j]<minn){
                minn=mind[j];
                k=j;
            }
         }
         vis[k]=true;
         res+=minn;
         for(int j=1;j<=n;j++){
            if(!vis[j]){
                mind[j]=min(mind[j],g[k][j]);
            }
         }
    }
    return res;
}

int main(){
    int u,v;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
        scanf("%lf%lf",&p[i].x,&p[i].y);
    }
    for(int i=1;i<=n;i++){
        for(int j=i+1;j<=n;j++){
            g[j][i]=g[i][j]=DST(p[i],p[j]);
        }
    }
    for(int i=0;i<m;i++){
        scanf("%d%d",&u,&v);
        g[u][v]=g[v][u]=0;
    }
    printf("%.2f
",prim());
}