zoukankan      html  css  js  c++  java
  • POJ 1751 Prim

    Highways
    Time Limit: 1000MS Memory Limit: 10000K
    Total Submissions: 8855 Accepted: 2467 Special Judge
    Description

    The island nation of Flatopia is perfectly flat. Unfortunately, Flatopia has a very poor system of public highways. The Flatopian government is aware of this problem and has already constructed a number of highways connecting some of the most important towns. However, there are still some towns that you can't reach via a highway. It is necessary to build more highways so that it will be possible to drive between any pair of towns without leaving the highway system.

    Flatopian towns are numbered from 1 to N and town i has a position given by the Cartesian coordinates (xi, yi). Each highway connects exaclty two towns. All highways (both the original ones and the ones that are to be built) follow straight lines, and thus their length is equal to Cartesian distance between towns. All highways can be used in both directions. Highways can freely cross each other, but a driver can only switch between highways at a town that is located at the end of both highways.

    The Flatopian government wants to minimize the cost of building new highways. However, they want to guarantee that every town is highway-reachable from every other town. Since Flatopia is so flat, the cost of a highway is always proportional to its length. Thus, the least expensive highway system will be the one that minimizes the total highways length.
    Input

    The input consists of two parts. The first part describes all towns in the country, and the second part describes all of the highways that have already been built.

    The first line of the input file contains a single integer N (1 <= N <= 750), representing the number of towns. The next N lines each contain two integers, xi and yi separated by a space. These values give the coordinates of ith town (for i from 1 to N). Coordinates will have an absolute value no greater than 10000. Every town has a unique location.

    The next line contains a single integer M (0 <= M <= 1000), representing the number of existing highways. The next M lines each contain a pair of integers separated by a space. These two integers give a pair of town numbers which are already connected by a highway. Each pair of towns is connected by at most one highway.
    Output

    Write to the output a single line for each new highway that should be built in order to connect all towns with minimal possible total length of new highways. Each highway should be presented by printing town numbers that this highway connects, separated by a space.

    If no new highways need to be built (all towns are already connected), then the output file should be created but it should be empty.
    Sample Input

    9
    1 5
    0 0
    3 2
    4 5
    5 1
    0 4
    5 2
    1 2
    5 3
    3
    1 3
    9 7
    1 2
    Sample Output

    1 6
    3 7
    4 9
    5 7
    8 3
    Source

    Northeastern Europe 1999


    <span style="color:#6600cc;">/**********************************************
    
          author   :    Grant Yuan
          time     :    2014.7.30
          algorithm:    Prim
          source   :    PKU 1751
    
    **********************************************/
    
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<algorithm>
    #include<vector>
    #include<stack>
    #include<queue>
    #include<map>
    #include<cmath>
    #define MAX 753
    #define INF 99999999
    
    using namespace std;
    
    struct node{int x,y;};
    int m,n;
    node b[MAX];
    int cost[MAX][MAX];
    bool used[MAX];
    int mincost[MAX];
    int pre[MAX];
    
    void prim()
    {
        for(int i=1;i<=n;i++)
        {
            mincost[i]=INF;
            used[i]=false;
        }
        memset(pre,0,sizeof(pre));
        mincost[1]=0;
        while(true){
            int v=-1;
            for(int i=1;i<=n;i++)
            {
    
                if(!used[i]&&(v==-1||mincost[i]<mincost[v]))
                   v=i;
            }
    
        if(v==-1) break;
        used[v]=true;
        for(int i=1;i<=n;i++)
        {
            if(cost[v][i]<mincost[i])
            {
                mincost[i]=cost[v][i];
                if(!used[i]) pre[i]=v;
            }
        }}
    }
    int main()
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&b[i].x,&b[i].y);
        }
        int d=1;
        for(int i=1;i<=n;i++)
          for(int j=i+1;j<=n;j++)
        {
           cost[i][j]=cost[j][i]=INF;cost[i][i]=INF;
           cost[i][j]=cost[j][i]=(b[i].x-b[j].x)* (b[i].x-b[j].x)+(b[i].y-b[j].y)* (b[i].y-b[j].y);
        }
        scanf("%d",&m);int aa,bb;
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d",&aa,&bb);
            cost[aa][bb]=cost[bb][aa]=0;
        }
    
        prim();
        for(int i=2;i<=n;i++)
        {
            if(cost[i][pre[i]]&&i!=pre[i])
                printf("%d %d
    ",i,pre[i]);
        }
    
    }
    
    
    
    
    
    
    
    
    
    
    
    </span>


  • 相关阅读:
    uni-app开发的应用(小程序,app,web等),使用Node+Koa2开发的后端程序接收上传文件的方法
    ES6学习笔记之数组的扩展
    ES6学习笔记之字符串的扩展
    UnhandledPromiseRejectionWarning: SequelizeConnectionError: Client does not support authentication protocol requested by server; consider upgrading MySQL client
    java之内存分布图
    java的构造方法
    java的局部变量和成员变量以及区别
    java类型转换详解(自动转换和强制转换)
    webpack入门宝典
    js函数之四大调用模式
  • 原文地址:https://www.cnblogs.com/codeyuan/p/4254470.html
Copyright © 2011-2022 走看看