洛谷P1550 [USACO08OCT]打井Watering Hole

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： 复制

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： 复制

9

说明

John等着用水，你只有1s时间！！！

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 310
using namespace std;
int n,fa[maxn],num;
struct node{
    int from,to,v;
    bool operator < (const node &a)const{
        return v<a.v;
    } 
}e[maxn*maxn];
int find(int x){
    if(x==fa[x])return x;
    return fa[x]=find(fa[x]);
}
int main(){
    scanf("%d",&n);
    int x;
    for(int i=0;i<=n;i++)fa[i]=i;
    for(int i=1;i<=n;i++){
        scanf("%d",&x);
        e[++num].from=0;e[num].to=i;e[num].v=x;
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            scanf("%d",&x);
            if(i>j)e[++num].from=i,e[num].to=j,e[num].v=x;
        }
    }
    sort(e+1,e+num+1);
    int ans=0;
    for(int i=1;i<=num;i++){
        int f1=find(e[i].from),f2=find(e[i].to);
        if(f1!=f2){
            ans+=e[i].v;
            fa[f1]=f2;
        }
    }
    printf("%d",ans);
    return 0;
}