bzoj1601[Usaco2008 Oct]灌水(最小生成树）

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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 需要为使所有农场都与有水的农场相连或拥有水井所需要的钱数。

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 #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 #1复制
9
说明/提示
John等着用水，你只有1s时间！！！

思路： 最小生成树

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <vector>using namespace std;typedef long long ll;
#define INF 0x3f3f3f3f
const ll maxn = 2e6 + 7;
struct Edge
{ll x,y;ll w;
}edges[maxn];ll fa[maxn];
ll findset(ll x)
{if(fa[x] == x)return x;return fa[x] = findset(fa[x]);
}ll cmp(Edge a,Edge b)
{return a.w < b.w;
}int main()
{ll n;scanf("%lld",&n);ll cnt = 0;for(ll i = 1;i <= n;i++){ll x;scanf("%lld",&x);edges[++cnt].x = 0;edges[cnt].y = i;edges[cnt].w = x;fa[i] = i;}for(ll i = 1;i <= n;i++){for(ll j = 1;j <= n;j++){ll tmp;scanf("%lld",&tmp);if(i == j)continue;edges[++cnt].x = i;edges[cnt].y = j;edges[cnt].w = tmp;}}sort(edges + 1,edges + 1 + cnt,cmp);ll ans = 0;ll num = 0;for(ll i = 1;i <= cnt;i++){ll ri = findset(edges[i].x),rj = findset(edges[i].y);if(ri != rj){fa[ri] = rj;ans += edges[i].w;num++;}else continue;if(num >= n)break;}printf("%lld\n",ans);return 0;
}