Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
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
3 3 1 2 3 3 1 2 4 4 1 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. 题意：给出n个点，找出一些等长的线段覆盖这些点。这些点仅仅能作为线段的端点，并且随意两条线段的相交长度不能大于0.求满足条件的线段的最大长度。
分析：通过分析能够得出。终于结果是相邻两点之间的长度，或者相邻两点之间长度的一半。由于最多仅仅有50个点，100个长度，所以仅仅需枚举这些长度。求出一个满足条件的最长线段就可以。
#include#include #include using namespace std;int main(){ double b[120], c[60]; int flag[60]; int n, i, j, T; scanf("%d",&T); while(T--) { scanf("%d",&n); for(i = 0; i < n; i++) scanf("%lf",&c[i]); sort(c, c+n); int m = 0; for(i = 1; i < n; i++) { b[m++] = c[i] - c[i-1]; b[m++] = (c[i] - c[i-1]) / 2; } sort(b, b+m); double ans; for(i = m - 1; i >= 0; i--) { memset(flag, 0, sizeof(flag)); flag[0] = 1; double tmp = b[i]; for(j = 1; j < n - 1; j++) { if(c[j] - tmp < c[j-1] && c[j] + tmp > c[j+1]) //往左往右都不行 break; if(c[j] - tmp >= c[j-1]) { if(flag[j-1] == 2) // 前一个往右 { if(c[j] - c[j-1] == tmp) flag[j] = 1; //两个点作为线段的两个端点 else if(c[j] - c[j-1] >= 2 * tmp) flag[j] = 1; //一个往左。一个往右 else if(c[j] + tmp <= c[j+1]) flag[j] = 2; //仅仅能往右 else break; } else flag[j] = 1; } else if(c[j] + tmp <= c[j+1]) flag[j] = 2; } if(j == n - 1) { ans = tmp; break; } } printf("%.3lf\n", double(ans)); } return 0;}