CF51C Three Base Stations

CF51C Three Base Stations - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

题目描述

The New Vasjuki village is stretched along the motorway and that's why every house on it is characterized by its shift relative to some fixed point — the x_{i}xi coordinate. The village consists of nn houses, the ii -th house is located in the point with coordinates of x_{i}xi .

TELE3, a cellular communication provider planned to locate three base stations so as to provide every house in the village with cellular communication. The base station having power dd located in the point tt provides with communication all the houses on the segment [t-d,t+d][t−d,t+d] (including boundaries).

To simplify the integration (and simply not to mix anything up) all the three stations are planned to possess the equal power of dd . Which minimal value of dd is enough to provide all the houses in the village with cellular communication.

输入格式

The first line contains an integer nn ( 1<=n<=2·10^{5}1<=n<=2⋅105 ) which represents the number of houses in the village. The second line contains the coordinates of houses — the sequence x_{1},x_{2},...,x_{n}x1,x2,...,xn of integer numbers ( 1<=x_{i}<=10^{9}1<=xi<=109 ). It is possible that two or more houses are located on one point. The coordinates are given in a arbitrary order.

输出格式

Print the required minimal power dd . In the second line print three numbers — the possible coordinates of the base stations' location. Print the coordinates with 6 digits after the decimal point. The positions of the stations can be any from 00 to 2·10^{9}2⋅109 inclusively. It is accepted for the base stations to have matching coordinates. If there are many solutions, print any of them.

题意翻译

一句话题意：给定一条直线上n个点（可能重合），现在你有3个基地，你可以选择在三个地方放置这三个基地，每个基地有一个属性值d，假设将一个基地放在t的位置，那么他可以辐射到[t-d,t+d]范围内的所有的点，现在要你求出最小的d，以及这三个基地放置的位置，保留6位小数

输入输出样例

输入 #1
4
1 2 3 4
输出 #1
0.500000
1.500000 2.500000 3.500000
输入 #2
3
10 20 30
输出 #2
0
10.000000 20.000000 30.000000
输入 #3
5
10003 10004 10001 10002 1
输出 #3
0.500000
1.000000 10001.500000 10003.500000

思路

找最小的属性值d，因为当d取最大值的时候可以满足，往小的方向取，不一定满足，所以我们需要用二分去寻找最小的满足条件的d。然后通过函数check去判断：第一个基站肯定是要放在第一个基站往右mid的距离，这样是最优的，然后在通过上一个确定的基站，去寻找第一个没有覆盖到的点，如果在找完所有的点之前，已经有了3个基站了，则这种d不正确，需要重新去找，把二分的左边界右移；如果满足条件，能够把所有的基站全部包括，则将结果进行记录，再把二分的右边界左移，直到完成二分，最终的结果就是最优的了。

代码实现

#include<bits/stdc++.h>
using namespace std;
int n;
double a[200005],l,r,d,b[3],c[3];
int check(double x)
{int cnt=0;b[cnt]=a[cnt]+x;//第一个基站设在a[0]+d(d在这里为x)时最优，因为能最大限度覆盖到其他点for(int i=1; i<n; i++){if(a[i]>b[cnt]+x)//下一个基站就建在上一个基站不能覆盖到的第一个点a[i]，然后放在a[i]+x的位置{if(cnt==2){return 0;}cnt++;b[cnt]=a[i]+x;}}for(int i=0; i<3; i++){c[i]=b[i];//记录基站位置}return 1;
}
int main()
{scanf("%d",&n);for(int i=0; i<n; i++){scanf("%lf",&a[i]);}sort(a,a+n);l=0,r=a[n-1];while(r-l>=1e-7) //精度误差{double mid=(l+r)/2;if(check(mid)==1){d=mid;r=mid-(1e-7);}else{l=mid+(1e-7);}}printf("%.6lf\n",d);printf("%.6lf %.6lf %.6lf",c[0],c[1],c[2]);return 0;
}

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CF51C Three Base Stations

题目描述

输入格式

输出格式

题意翻译

输入输出样例

思路

代码实现

CF51C Three Base Stations相关推荐

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