[题解] 洛谷 P1550 USACO 2008 OCT 打井Watering Hole (kruskal最小生成树)

- 传送门 -

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

# P1550 [USACO08OCT]打井Watering Hole

时空限制 1s / 128MB

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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！！！
　

- 题意 -

　见题面.
　

- 思路 -

　一开始想错了, 蓝瘦...
　建一个虚点, 从每个点向其连一条边权=点权的边, 和虚点相连的点表示直接供水.
　从图中取出 n 条边即可(至少要有一条边与虚点相连).
　
　细节见代码.
　
　PS:
　没有权限号的人一开始去找bzoj1601是生气的:)
　

- 代码 -

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int N = 305;

struct edge {
	int x, y, v;
}W[N*N];

int C[N], D[N][N];
int F[N];
int n, ans, tot, sz, cnt;

bool cmp(edge a, edge b) { return a.v < b.v; }

int find(int x) { return F[x] == x ? x : F[x] = find(F[x]); }

void kruskal() {
	for (int i = 1; i <= sz; ++i) {
		int xf = find(W[i].x), yf = find(W[i].y);
		if (xf == yf) continue;
		F[xf] = yf;
		ans += W[i].v;
		cnt ++;
		if (cnt == n) break;
	}
}

void add(int a, int b, int v) {
	W[++sz].x = a;
	W[sz].y = b;
	W[sz].v = v;
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%d", &C[i]);
		F[i] = i;
		add(i, 0, C[i]);
	}
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= n; ++j) {
			scanf("%d", &D[i][j]);
			if (j > i)
				add(i, j, D[i][j]);
		}
	sort(W + 1, W + sz + 1, cmp);
	kruskal();
	printf("%d\n", ans);
	return 0;
}