luogu P1550 [USACO08OCT]打井Watering Hole

题目背景

John的农场缺水了!!!

题目描述

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.

POINTS: 400

农民John 决定将水引入到他的n(1<=n<=300)个牧场。他准备通过挖若

干井,并在各块田中修筑水道来连通各块田地以供水。在第i 号田中挖一口井需要花费W_i(1<=W_i<=100,000)元。连接i 号田与j 号田需要P_ij (1 <= P_ij <= 100,000 , P_ji=P_ij)元。

请求出农民John 需要为连通整个牧场的每一块田地所需要的钱数。

输入输出格式

输入格式:

第1 行为一个整数n。

第2 到n+1 行每行一个整数,从上到下分别为W_1 到W_n。

第n+2 到2n+1 行为一个矩阵,表示需要的经费(P_ij)。

输出格式:

只有一行,为一个整数,表示所需要的钱数。

输入输出样例

输入样例#1: 
4
5
4
4
3
0 2 2 2
2 0 3 3
2 3 0 4
2 3 4 0
输出样例#1: 
9

说明

John等着用水,你只有1s时间!!!

开始给dis赋值打井花费来比较打井和修水道哪种更优

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 using namespace std;
 5 int f[310][310],dis[310];
 6 bool vis[310]; 
 7 int main()
 8 {
 9     memset(f,0x3f,sizeof(f));
10     memset(dis,0x3f,sizeof(dis));
11     int n;
12     scanf("%d",&n);
13     for(int i=1;i<=n;++i)
14         scanf("%d",&dis[i]);
15     for(int i=1;i<=n;++i)
16         for(int j=1;j<=n;++j)
17             scanf("%d",&f[i][j]);
18     for(int i=1;i<=n;++i)
19     {
20         int k=0;
21         for(int j=1;j<=n;++j)
22             if(!vis[j]&&dis[j]<dis[k]) 
23                 k=j;
24         vis[k]=1;
25         for(int j=1;j<=n;++j)
26             if(j!=k&&!vis[j]&&dis[j]>f[k][j])
27                 dis[j]=f[k][j];
28     }
29     int ans=0;
30     for(int i=1;i<=n;++i)
31         ans+=dis[i]; 
32     printf("%d",ans);
33     return 0;
34 }

猜你喜欢

转载自www.cnblogs.com/axma/p/9356912.html