BestCoder Round #4 之 Miaomiao's Geometry（2014/8/10）
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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
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.
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.
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：
#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;
}
转载于:https://www.cnblogs.com/yspworld/p/3903330.html
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