洛谷 P2212 [USACO14MAR]Watering the Fields S 题解

2021-08-03

20:31:13

链接：

https://www.luogu.com.cn/problem/P2212

题目详情：

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,cn,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复制

3 11
0 2
5 0
4 3

输出 #1复制

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

对于 100% 的数据，1≤n≤2000，0≤xi​,yi​≤1000，1≤c≤10^6。

题目分析：

这是一道prim算法题，我们可以利用prim算法的模板，加上题目的限制条件两点距离不得小于c，就可以解出该题。

话不多说直接上代码吧

#include<iostream>
#include<cmath>
using namespace std;
const int N=2005;
const int INF=1e8;
int n,c;
struct Node{
    int x,y;
}node[N];
int g[N][N];//每个点之间的距离
int dist[N];
bool st[N];
int distance(int x1,int y1,int x2,int y2){
    int t=pow((x1-x2),2)+pow((y1-y2),2);
    return t;
}

int Prim(){
    int res=0;
    for(int i=0;i<n-1;i++){
        for(int j=i+1;j<n;j++){
            int x1=node[i].x,y1=node[i].y,x2=node[j].x,y2=node[j].y;
            int t=distance(x1,y1,x2,y2);
            if(t>=c)g[i][j]=g[j][i]=t;
            else g[i][j]=g[j][i]=INF;
        }
    }
    for(int i=0;i<n;i++){
        int t=-1;
        for(int j=0;j<n;j++){
            if(!st[j]&&(t==-1||dist[t]>dist[j]))t=j;
        }
        if(i&&dist[t]==INF)return -1;
        if(i)res+=dist[t];
        st[t]=true;
        for(int j=1;j<n;j++)
            dist[j]=min(dist[j],g[t][j]);
    }
    return res;
}

int main()
{
    cin>>n>>c;
    for(int i=0;i<n;i++){
        int x,y;
        cin>>x>>y;
        node[i]={x,y};
        dist[i]=INF;
        g[i][i]=INF;
    }
    int res=Prim();
    cout<<res;
    return 0;
}

　

2021-08-03

20:43:34