[贪心]T-net

最新推荐文章于 2025-02-19 15:02:42 发布

转载最新推荐文章于 2025-02-19 15:02:42 发布 · 112 阅读

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原文链接：http://www.cnblogs.com/lllxq/p/10691535.html

本文探讨了T-net电信公司如何在城市中安装基站，使用两种不同传输半径的天线，以确保整个网络连接的同时，最小化总成本。文章提供了一种算法解决方案，通过分析基站位置和天线覆盖范围，实现网络的最优化构建。

题目描述

T-net which is a new telecommunications company, plans to install its base stations in the city. The places where base stations must be installed have been already specified. T-net has two types of antennas to be used in the base stations: (i)antennas with transmission radius a, and (ii) antennas with transmission radius b. Two antennas can communicate with each other if and only if both are inside the coverage area of each other. Antenna with smaller transmission radius of course is cheaper. T-net plans to minimize its cost while keeping the whole network connected. Precisely, T-net plans to
minimize its cost which is the sum of the transmission radii of all antennas. Interestingly, all base-station places are on a line. Help T-net construct a connected network with the minimum cost.

输入

The first line of the input contains three positive integers n, a and b (1 ⩽ n ⩽ 105 and 1 ⩽ a, b ⩽ 105 ) where n is the number of base stations, and a and b are radii as defined in the problem statement. The second line contains n distinct coordinates of base stations on the line with respect to an origin on the line. All coordinates are positive integers not more than 105 .

输出

If it is possible to construct a connected network, print the minimum cost in the output. Otherwise, print -1 .

样例输入

复制样例数据

3 1 3
1 4 3

样例输出

#include<bits/stdc++.h>
using namespace std;

int s[100005];
int d[100005];

int main()
{
    int n,a,b;
    scanf("%d%d%d",&n,&a,&b);
    if(a>b) swap(a,b);
    for(int i=1;i<=n;i++) scanf("%d",&s[i]);
    sort(s+1,s+1+n);
    for(int i=2;i<=n;i++) if(s[i]-s[i-1]>b) return 0*puts("-1");
    s[0]=s[1],s[n+1]=s[n];
    for(int i=1;i<=n;i++){
        if(d[i]==a) continue;
        if(!d[i]){
            if(s[i]-s[i-1]<=a&&s[i+1]-s[i]<=a) {d[i]=a;continue;}
            else d[i]=b;
        }
        int j=i;
        while(s[i]+b>=s[j]&&j<=n) j++;
        j--;
        int flag=0;
        for(int k=i+1;k<=j-1;j++){
            if(s[k]-s[k-1]>a||s[k+1]-s[k]>a){
                flag=1;
                break;
            }
        }
        if(!flag) continue;
        d[j]=b;
        for(int k=i+1;k<=j-1;k++){
            if(s[k]-s[k-1]<=a) d[k]=a;
            else break;
        }
        for(int k=j-1;k>=i+1;k--){
            if(s[k+1]-s[k]<=a) d[k]=a;
            else break;
        }
    }
    long long ans=0;
    for(int i=1;i<=n;i++) ans+=d[i];
    cout<<ans<<endl;
}