分享一个不错的题目“典型的最小生成树的应用”

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本文探讨了如何利用最小生成树算法解决农民John为多个牧场提供水源的问题，通过合理选择建设水井或铺设管道，最小化总成本。

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Watering Hole

My Tags	(Edit)

	Source : USACO 2008 Open
	Time limit : 1 sec		Memory limit : 64 M

Submitted : 71, Accepted : 48

Farmer John has decided to bring water to his N (1 <= N <= 300) pastures which are conveniently numbered 1..N. He may bring water to a pasture either by building a well in that pasture or connecting the pasture via a pipe to another pasture which already has water.

Digging a well in pasture i costs W_i (1 <= W_i <= 100,000). Connecting pastures i and j with a pipe costs P_ij (1 <= P_ij <= 100,000; P_ij = P_ji; P_ii=0).

Determine the minimum amount Farmer John will have to pay to water all of his pastures.

INPUT FORMAT

* Line 1: A single integer: N

* Lines 2..N + 1: Line i+1 contains a single integer: W_i

* Lines N+2..2N+1: Line N+1+i contains N space-separated integers; the j-th integer is P_ij

SAMPLE INPUT

INPUT DETAILS

There are four pastures. It costs 5 to build a well in pasture 1, 4 in pastures 2 and 3, 3 in pasture 4. Pipes cost 2, 3, and 4 depending on which pastures they connect.

OUTPUT FORMAT

* Line 1: A single line with a single integer that is the minimum cost of providing all the pastures with water.

SAMPLE OUTPUT

OUTPUT DETAILS

Farmer John may build a well in the fourth pasture and connect each pasture to the first, which costs 3 + 2 + 2 + 2 = 9.

题目分析：

咋一看上去似乎无解，如果是DP的话，每个pastures被浇水的选择有两种:一种是打井，一种是从其他的pastures连接水管，但是状态呢？这个状态要能够记录一个集合里那些是被浇过水的哪些待浇水的，300个定点状态达到2^300，无解; 我还想过用最小费用最大流的方式，把打井抽象出来一个节点然后该节点向每个pastures连接边流量为n，费用为对应pastures打井的费用，然后新修一个虚拟节点汇点，流量为1费用为0，这样做结果显然不对的。

比较优美的解法是最小生成树，把打井抽象出一个节点0，其他的pastures分别用1 ~ n表示。

代码都不解释了，一般提到最小生成树都可以想出来了

#include<iostream>

using namespace std;

int g[310][310];
int dist[310];
bool vis[310];

int prime(int n)
{
    memset(vis,0,sizeof(vis));
    int ans = 0;
    for(int i=0;i<=n;i++)
    {
       dist[i] = g[0][i];
       vis[i] = 0;
    }
    vis[0] = 1;
    
    for(int i=0;i<n;i++)
    {
        int k = -1,Min = 99999999;
        for(int j=0;j<=n;j++)
          if(!vis[j] && Min > dist[j]) {
               k = j;
               Min = dist[j];
          }
        vis[k] = 1;
        ans += Min;
        for(int j=0;j<=n;j++)
           if(!vis[j]&&dist[j] > g[k][j])
              dist[j] = g[k][j];
    }
    
    return ans;
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<=n;i++)
    for(int j=1;j<=n;j++)
      scanf("%d",&g[i][j]);
      
    for(int i=1;i<=n;i++)
       g[i][0] = g[0][i];
       
    /*for(int i=0;i<=n;i++)
    {
       for(int j=0;j<=n;j++)
         cout << g[i][j] << " ";
       cout << endl;
    }*/
    
    printf("%d\n",prime(n));
   
    return 0;
}
/*
4
5
4
4
3
0 2 2 2
2 0 3 3
2 3 0 4
2 3 4 0
*/