題目背景

John的農場缺水了！！！

題目描述

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

農民John 決定將水引入到他的n(1<=n<=300)個牧場。他準備通過挖若

乾井，並在各塊田中修築水道來連通各塊田地以供水。在第i 號田中挖一口井需要花費W_i(1<=W_i<=100,000)元。連線i 號田與j 號田需要P_ij (1 <= P_ij <= 100,000 , P_ji=P_ij)元。

請求出農民John 需要為使所有農場都與有水的農場相連或擁有水井所需要的錢數。

輸入輸出格式

輸入格式：

第1 行為一個整數n。

第2 到n+1 行每行一個整數，從上到下分別為W_1 到W_n。

第n+2 到2n+1 行為一個矩陣，表示需要的經費（P_ij）。

輸出格式：

只有一行，為一個整數，表示所需要的錢數。

輸入輸出樣例

輸入樣例#1： 

4
5
4
4
3
0 2 2 2
2 0 3 3
2 3 0 4
2 3 4 0

輸出樣例#1： 

9

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int n,m;
long long ans;
struct node{
	int x;
	int y;
	int z;
}a[100005];
int s[100055];
int fa[100005];
int cnt=0;
bool cmp(const node &x,const node &y)
{
	return x.z<y.z;
}
int find(int x)
{
	if(fa[x]==x)
	{
		return x;
	}
	return fa[x]=find(fa[x]);
}
void kkk()
{
	int f1;
	int f2;
	int k=0;
	for(int i=1;i<=n;i++)
	fa[i]=i;
	for(int i=1;i<=cnt;i++)
	{
		f1=find(a[i].x);
		f2=find(a[i].y);
		if(f1!=f2)
		{
			ans=ans+a[i].z;
			fa[f1]=f2;
			k++;
			if(k==n-1)
			break;
		}
	}
} 
int main(){
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&s[i]);
	}
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=n;j++)
		{
			scanf("%d",&a[++cnt].z);
			a[cnt].x=i;
			a[cnt].y=j;
		}
		a[++cnt].x=i;
		a[cnt].y=n+1;
		a[cnt].z=s[i];
	}
	n++;
	sort(a+1,a+1+cnt,cmp);
	kkk();
	cout<<ans<<endl;
}