題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=4932

Miaomiao's Geometry

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 410????Accepted Submission(s): 147

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 HintFor 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. Source BestCoder Round #4

題意：

求最大可以覆蓋全部所給的點的區間長度（所給的點必須處于區間兩端）。

思路：

? ? ? ? 答案一定是相鄰點之間的差值或者是相鄰點之間的差值除以2，那么把這些可能的答案先算出來。然后依次從最大的開始枚舉進行驗證就可以。

代碼例如以下：

#include <cstdio> #include <algorithm> #include <cstring> using namespace std; const int MAXN = 147; int f[MAXN];//記錄線段方向 double p[MAXN]; double d[MAXN];//相鄰斷點的差值 int n; void init() {memset(p,0,sizeof(p));memset(f,0,sizeof(f));memset(d,0,sizeof(d)); }bool Judge(double tt) {int i;for(i = 1; i < n-1; i++){if(p[i] - tt < p[i-1] && p[i] + tt > p[i+1])break;//不管向左還是向右均為不符合if(p[i] - tt >= p[i-1])//向左察看{if(f[i-1] == 2)//假設前一個是向右的{if(p[i] - p[i-1] == tt)f[i] = 1;//兩個點作為線段的兩個端點else if(p[i] - p[i-1] >= 2*tt)//一個向左一個向右{f[i] = 1;}else if(p[i] + tt <= p[i+1]){f[i] = 2;//僅僅能向右}elsereturn false;}elsef[i] = 1;}else if(p[i] + tt <= p[i+1])f[i] = 2;}if(i == n-1)//所有符合return true;return false; } int main() {int t;scanf("%d",&t);while(t--){init();scanf("%d",&n);for(int i = 0; i < n; i++){scanf("%lf",&p[i]);}sort(p,p+n);int cont = 0;for(int i = 1; i < n; i++){d[cont++] = p[i] - p[i-1];d[cont++] = (p[i] - p[i-1])/2.0;}sort(d,d+cont);double ans = 0;for(int i = cont-1; i >= 0; i--){memset(f,0,sizeof(f));f[0] = 1; //開始肯定是讓線段向左if(Judge(d[i])){ans = d[i];break;}}printf("%.3lf\n",ans);}return 0; }

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