题目描述

Due to a lack of rain, Farmer John wants to build an irrigation system to

send water between his N fields (1 <= N <= 2000).

Each field i is described by a distinct point (xi, yi) in the 2D plane,

with 0 <= xi, yi <= 1000. The cost of building a water pipe between two

fields i and j is equal to the squared Euclidean distance between them:

(xi - xj)^2 + (yi - yj)^2

FJ would like to build a minimum-cost system of pipes so that all of his

fields are linked together -- so that water in any field can follow a

sequence of pipes to reach any other field.

Unfortunately, the contractor who is helping FJ install his irrigation

system refuses to install any pipe unless its cost (squared Euclidean

length) is at least C (1 <= C <= 1,000,000).

Please help FJ compute the minimum amount he will need pay to connect all

his fields with a network of pipes.

给定 n 个点，第 i 个点的坐标为 (xi,yi)，如果想连通第 i 个点与第 j 个点，需要耗费的代价为两点的距离。第 i 个点与第 j 个点之间的距离使用欧几里得距离进行计算，即：

(xi−xj)2+(yi−yj)2

我们规定耗费代价小于 c 的两点无法连通，求使得每两点都能连通下的最小代价，如果无法连通输出 -1。

输入格式

* Line 1: The integers N and C.

* Lines 2..1+N: Line i+1 contains the integers xi and yi.

第一行两个整数 n,c 代表点数与想要连通代价不能少于的一个数。
接下来 n 行每行两个整数 xi,yi 描述第 i 个点。

输出格式

* Line 1: The minimum cost of a network of pipes connecting the

fields, or -1 if no such network can be built.

一行一个整数代表使得每两点都能连通下的最小代价，如果无法连通输出 -1。

输入输出样例

输入 #1
3 11
0 2
5 0
4 3
输出 #1
46
说明/提示

INPUT DETAILS:

There are 3 fields, at locations (0,2), (5,0), and (4,3). The contractor

will only install pipes of cost at least 11.

OUTPUT DETAILS:

FJ cannot build a pipe between the fields at (4,3) and (5,0), since its

cost would be only 10. He therefore builds a pipe between (0,2) and (5,0)

at cost 29, and a pipe between (0,2) and (4,3) at cost 17.

Source: USACO 2014 March Contest, Silver

数据规模与约定

说明

Translated by 一只书虫仔。

思路：

建图跑最小生成树

AC代码如下：

#include <bits/stdc++.h>
#define buff                     \ios::sync_with_stdio(false); \cin.tie(0);                  \cout.tie(0)
using namespace std;
int n,c;
const int N = 2009;
double g[N][N];
bool st[N];
double dist[N];
struct point
{double x,y;
}s[N];
void add(int a, int b)
{double d = (s[a].x - s[b].x) * (s[a].x - s[b].x) + (s[a].y - s[b].y) * (s[a].y - s[b].y);if(d>=c)g[a][b] = d;elseg[a][b] = 0x3f3f3f3f;
}
void init()
{for(int i = 1;i<=n;i++)dist[i] = 0x3f3f3f3f;
}
int prim()
{init();double res = 0;for(int i = 0;i<n;i++){int t = -1;for(int j = 1;j<=n;j++){if(!st[j]&&(t==-1||dist[t]>dist[j]))t = j;}if(i&&dist[t]==0x3f3f3f3f)return -1;if(i)res += dist[t];st[t] = 1;for(int j = 1;j<=n;j++)dist[j] = min(dist[j],g[t][j]);}return res;
}
int main()
{buff;cin>>n>>c;for(int i = 1;i<=n;i++){cin>>s[i].x>>s[i].y;}for(int i = 1;i<=n;i++)for(int j = 1;j<=n;j++){if(i!=j)add(i,j);}int ans = prim();cout<<ans<<endl;
}

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P2212 [USACO14MAR]Watering the Fields S（最小生成树）

题目描述

输入格式

输出格式

输入输出样例

说明/提示

数据规模与约定

说明

思路：

AC代码如下：

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