hdu 4932 Miaomiao's Geometry（暴力枚舉）

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Miaomiao‘s Geometry                                                                             Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem DescriptionThere 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. 
InputThere 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. 
OutputFor 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

1.0002.0008.000HintFor 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. 

 題意：給出n個點，找出一些等長的線段覆蓋這些點，這些點只能作為線段的端點，而且任意兩條線段的相交長度不能大於0.求滿足條件的線段的最大長度。分析：通過分析可以得出，最終結果是相鄰兩點之間的長度，或者相鄰兩點之間長度的一半。因為最多隻有50個點，100個長度，所以只需枚舉這些長度，求出一個滿足條件的最長線段即可。

#include<cstdio>#include<cstring>#include<algorithm>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) // 前一個往右                    {                        if(c[j] - c[j-1] == tmp) flag[j] = 1;  //兩個點作為線段的兩個端點                        else if(c[j] - c[j-1] >= 2 * tmp) flag[j] = 1; //一個往左，一個往右                        else if(c[j] + tmp <= c[j+1]) flag[j] = 2; //只能往右                        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;}