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.

翻译

农民John 决定将水引入到他的边长为n(1<=n<=300)的牧场。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！！！

用最少的花费把所有点都连起来，最小生成树……井的话就看做一个超级源点0号点，这就成了求 0 - n 这 n + 1 个点的最小生成树。

代码如下

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int size = 200010;
int n;
int num[size];
struct dc
{
int f,t,d;
}l[size];
int tot = 1;
void build(int f,int t,int d)
{
l[tot].f = f;
l[tot].t = t;
l[tot].d = d;
tot ++;
}
int f[size];
int find(int x)
{
if(f[x] == x)
return x;
return f[x] = find(f[x]);
}
bool cmp(dc a,dc b)
{
return a.d < b.d;
}
int main()
{
scanf("%d",&n);
for(int i = 1 ; i <= n ; i ++)
scanf("%d",&num[i]) , f[i] = i;
for(int i = 1 ; i <= n ; i ++)
for(int j = 1 ; j <= n ; j ++)
{
int d;
scanf("%d",&d);
build(i,j,d);
}
for(int i = 1 ; i <= n ; i ++)
build(0,i,num[i]);
sort(l+1,l+tot,cmp);
int ans = 0;
for(int i = 1 ; i < tot ; i ++)
{
int ff = l[i].f , ft = l[i].t;
if(find(ff) != find(ft))
{
f[find(ff)] = find(ft);
ans += l[i].d;
}
}
printf("%d\n",ans);
return 0;
}

传送门：https://www.luogu.org/problem/show?pid=1550