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There are N point on X-axis . Miaomiao would like to cover them ALL by using segments with same length.
There are 2 limits:
1.A point is convered if there is a segments T , the point is the left end or the right end of T.
2.The length of the intersection of any two segments equals zero.
For example , point 2 is convered by [2 , 4] and not convered by [1 , 3]. [1 , 2] and [2 , 3] are legal segments , [1 , 2] and [3 , 4] are legal segments , but [1 , 3] and [2 , 4] are not (the length of intersection doesn‘t equals zero), [1 , 3] and [3 , 4] are not(not the same length).
Miaomiao wants to maximum the length of segements , please tell her the maximum length of segments.
For your information , the point can‘t coincidently at the same position.
/*
大意：在n个点之间放线段，问最大多少
去重
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int n, T;
const double inf= 1e9 + 100;
const int MAX = 55;
double a[MAX];
bool check(double x)
{
double cur = a[1];
for(int i = 2; i <= n ; i++){
if(cur > a[i]) return false;
else if(cur == a[i]) continue;
if (a[i] - cur >= x){
cur = a[i];
}
else cur = a[i] + x;
}
return true;
}
int main()
{
scanf("%d", &T);
while(T--){
scanf("%d", &n);
for(int i = 1; i <= n ;i++)
scanf("%lf", &a[i]);
sort(a + 1, a + n + 1);
double l = 0, r = inf;
for(int i = 1; i <= 100; i++){
double mid = (l + r) / 2 ;
if(check(mid)){
l = mid ;
}
else r = mid ;
}
for(int i = 1; i < n; i++){
if(check(a[i+1] - a[i]) && (a[i+1] - a[i] > l))
l = a[i+1] - a[i];
}
printf("%.3f\n", l);
}
return 0;
}
HDU4932——二分——Miaomiao's Geometry
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原文地址：http://www.cnblogs.com/zero-begin/p/4694176.html
