zoukankan      html  css  js  c++  java
  • poj 3625 (最小生成树算法)

    Building Roads
    Time Limit: 1000MS Memory Limit: 65536K
    Total Submissions: 12203 Accepted: 3448

    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 (XiYi) on the plane (0 ≤ X≤ 1,000,000; 0 ≤ Y≤ 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: Xand Y
    * 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

    #include<iostream>
    #include<stdio.h>
    #include<cmath>
    #define MAX_V 1005
    using namespace std;


    double cost[MAX_V][MAX_V];
    double mincost[MAX_V];
    bool used[MAX_V];
    double X[MAX_V];double Y[MAX_V];
    int V;
    double min(double d1,double d2)
    {
        return d1<d2?d1:d2;
    }
    double prim()
    {
        for(int i=0;i<V;++i)
        {
            mincost[i]=1000000000.0;
            used[i]=false;
        }
        mincost[0]=0;
        double res=0;
        while(true)
        {
            int v=-1;
            for(int u=0;u<V;u++)
            {
                if(!used[u]&&(v==-1||mincost[u]<mincost[v]))v=u;
            }
            if(v==-1)break;
            used[v]=true;
            res+=mincost[v];
            for(int u=0;u<V;u++)
            {
                mincost[u]=min(mincost[u],cost[v][u]);
            }
        }
        return res;
    }


    double  dis(int i,int j)
    {
        return sqrt((X[i]-X[j])*(X[i]-X[j])+(Y[i]-Y[j])*(Y[i]-Y[j]));
    }


    int main()
    {
        freopen("in.txt","r",stdin);
        int M;
        cin>>V>>M;


        for(int i=0;i<V;i++)
        {
            cin>>X[i]>>Y[i];
        }


        for(int i=0;i<V;i++)
        {
            for(int j=i+1;j<V;j++)
            {
                cost[i][j]=cost[j][i]=dis(i,j);
            }
        }
        for(int i=0;i<V;i++)cost[i][i]=1000000000.0;
        int temp1,temp2;
        while(M--)
        {
            cin>>temp1>>temp2;
            //已经修建好的路则应理解为花费cost值为0
            cost[temp1-1][temp2-1]=cost[temp2-1][temp1-1]=0;
        }
        double ans=prim();
        printf("%.2f ",ans);
        return 0;
    }

  • 相关阅读:
    SpringBoot学习:整合shiro(身份认证和权限认证),使用EhCache缓存
    帝国备份王出错
    spring boot整合mybatis+mybatis-plus
    Druid连接池简介和配置
    thinkphp生成的验证码不显示问题解决
    分布式文件系统-FastDFS
    Spring Security OAuth2 Demo
    spring cloud-给Eureka Server加上安全的用户认证
    spring cloud 报错Error creating bean with name 'hystrixCommandAspect' ,解决方案
    分布式唯一ID极简教程
  • 原文地址:https://www.cnblogs.com/linruier/p/9485184.html
Copyright © 2011-2022 走看看