首页 > 代码库 > BestCoder Round #4 之 Miaomiao's Geometry（2014/8/10）

BestCoder Round #4 之 Miaomiao's Geometry（2014/8/10）

2024-07-16 19:54:46   224人阅读

                         Miaomiao‘s Geometry


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 10    Accepted Submission(s): 3


Problem Description
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.
 
Input
There are several test cases.
There is a number T ( T <= 50 ) on the first line which shows the number of test cases.
For each test cases , there is a number N ( 3 <= N <= 50 ) on the first line.
On the second line , there are N integers Ai (-1e9 <= Ai <= 1e9) shows the position of each point.
 
Output
For each test cases , output a real number shows the answser. Please output three digit after the decimal point.
 
Sample Input
331 2 331 2 441 9 100 10
 
Sample Output
1.000
2.000
8.000
Hint
For the first sample , a legal answer is [1,2] [2,3] so the length is 1. For the second sample , a legal answer is [-1,1] [2,4] so the answer is 2. For the thired sample , a legal answer is [-7,1] , [1,9] , [10,18] , [100,108] so the answer is 8.
题目大意：
类似于“区间覆盖”的问题！
在每组输入n个数，这n个数字是不同的，也就意味着，每两个值之间有差值。
题目要求用不计数的相等长度线段去覆盖这些点，并且呢，这些点必须在所在小线段的端点上。
举例说明（上面的第三组数据）：
a数组： 1 9 100 10       先排一下序，
--->1 9 10 100 要覆盖这些点，并且这些点都必须在端点上，并且用于
覆盖的任何两条线段都不能有重复的部分。
b数组 8 1 90 ----> 相邻两数之间的间距，给b数组也排序。
把b数组的值遍历于a数组：思路是：a数组的第一个值和最后一个值可以往外扩展，则不必考虑了。
现在考虑a数组中间部分的数值，对于每个a[i](i=1--->=n-2)要满足:(a[i]+b[j]<a[i+1] || a[i]-b[j]>a[i-1] ),如果当前的b[j]满足所有的a[i]，则说明b[j]可行，但是我们要找一个最大的b[j].
循环遍历一下b数组找到就行了。
 
Accepted的代码如下：（可供参考)
#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int main(){	int t;	int n;	int i, j;	int a[60];    int b[60], e;	scanf("%d", &t);	while(t--)	{		scanf("%d", &n);		for(i=0; i<n; i++)		{			scanf("%d", &a[i] );		}		sort(a, a+n);				e=0;		for(i=1; i<n; i++)		{			b[e++] = a[i] - a[i-1] ; 		}		sort(b, b+n-1 );				int max=-1;		int flag;				for(i=0; i<n-1; i++)		{			flag=1; //初始化每个间距标记			for(j=1; j<n-1; j++)			{					if(a[j]-b[i]<a[j-1] && a[j]+b[i]>a[j+1] )				{					flag=0;					break;				}			}			if(flag==1)			{				if(b[i] >max )				{					max = b[i] ;				}			}		}		printf("%d", max );		printf(".000\n");	}	return 0;}

 

 
 


联系
我们