zoukankan      html  css  js  c++  java
  • POJ 1258 Agri-Net( 最小生成树-Prim)

    Farmer John has been elected mayor of his town! One of his campaign promises was to bring internet connectivity to all farms in the area. He needs your help, of course.
    Farmer John ordered a high speed connection for his farm and is going to share his connectivity with the other farmers. To minimize cost, he wants to lay the minimum amount of optical fiber to connect his farm to all the other farms. 
    Given a list of how much fiber it takes to connect each pair of farms, you must find the minimum amount of fiber needed to connect them all together. Each farm must connect to some other farm such that a packet can flow from any one farm to any other farm. 
    The distance between any two farms will not exceed 100,000. 

    Input

    The input includes several cases. For each case, the first line contains the number of farms, N (3 <= N <= 100). The following lines contain the N x N conectivity matrix, where each element shows the distance from on farm to another. Logically, they are N lines of N space-separated integers. Physically, they are limited in length to 80 characters, so some lines continue onto others. Of course, the diagonal will be 0, since the distance from farm i to itself is not interesting for this problem.

    Output

    For each case, output a single integer length that is the sum of the minimum length of fiber required to connect the entire set of farms.

    Sample Input

    4
    0 4 9 21
    4 0 8 17
    9 8 0 16
    21 17 16 0
    

    Sample Output

    28
    题解:
      没什么技巧,直接上模板。
    #include<iostream>
    #include<algorithm>
    #include<cstdio>
    #include<cmath>
    using namespace std;
    const int maxn=105,INF=1e9;
    int cost[maxn][maxn];
    int mincost[maxn];
    bool used[maxn];
    int n;
    int prim()
    {
        for(int i=0;i<n;i++)
        {
            mincost[i]=INF;
            used[i]=false;
        }
        mincost[0]=0;
        int res=0;
        while(666)
        {
            int v=-1;
            for(int u=0;u<n;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<n;u++)
                mincost[u]=min(mincost[u],cost[v][u]);
        }
        return res;
    }
    int main()
    {
        while(cin>>n)
        {
            for(int i=0;i<n;i++)
                for(int j=0;j<n;j++)
                    scanf("%d",&cost[i][j]);
            cout<<prim()<<endl;
        }
        return 0;
    }
    
    
  • 相关阅读:
    ES6 一些新特性的总结
    前端模块化开发的规范:AMD与CDM
    webpack与grunt/glub 的比较
    前端总结(一)
    前端性能的优化
    Typescript 常见写法
    显示模式------行内元素、块元素,行内块元素
    浏览器前缀及内核
    BFC规范
    数据库习题练习
  • 原文地址:https://www.cnblogs.com/orion7/p/7373220.html
Copyright © 2011-2022 走看看