POJ 1258 Agri-Net (最小生成树+Prim)

Agri-Net

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 39820		Accepted: 16192

Description

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

Source

USACO 102

代码：(新手 勿喷)

Slyar:简介一下题意。农民要建立互联网络，目的使村庄里全部的农民连上网，而且总费用最小。多组数据。每组数据给出一个n，然后给出n * n大小的无向图的邻接矩阵表示，值表示边权。

要求输出最小生成树的权值和。

能够用Kruskal算法解决该题。用并查集检查待增加生成树的两边是否会构成回路，高速排序按权值排列边。

这里有一个优化:由于是无向图，所以矩阵是对称的，因此我们仅仅保存上三角矩阵就可以。这样到最后k的值就是边数。由循环n*n次缩减到循环k次...

只是我用的是Prim......

全是模板啦。我也不怎么会解释，尽量啦。

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
#define M 101
#define INF 1000001
int map[M][M];
int n;
int Prim()
{
    int s=1;
    int m=1;
    bool cmp[M]={0};   //标记这个点是否在子图内。
    cmp[s]=1;          //从第一个点開始。

int min_w;         //用来找当前的最小权边。

int prim_w=0;      //用来存线路的总长。
    int point;         //这是用来存那个即将入图的那个点的。
    int low_dis[M];    //这个用来存子图到这个点的最短距离。（关键）
    for(int i=1;i<=n;i++)
        low_dis[i]=INF;  //初始化一定要大。
    while(1)
    {
        if(m==n) break; //假设每一个点都入图，已经将全部点连好了。

min_w=INF;
        for(int i=2;i<=n;i++)
        {
            if(!cmp[i] && low_dis[i]>map[s][i])
                low_dis[i]=map[s][i];          //假设子图有更短的才替换（关键）
            if(!cmp[i] && min_w>low_dis[i])
            {
                min_w=low_dis[i];
                point=i;                      //这就是找那个最小权边。并标记它的下标。
            }
        }
        s=point;
        cmp[s]=1;                           //这个点入图，加上权边值。
        prim_w+=min_w;

        m++;  //子图中的点数+1.
      }
    return prim_w;
}
int main()
{
    int i,j;
    while(cin>>n)
    {
        for(i=1;i<=n;i++)
        for(j=1;j<=n;j++)
        cin>>map[i][j];
        cout<<Prim()<<endl;
    }
    return 0;
}

我也没办法。

。。模板就这样。。

。