【CUGBACM15级BC第四场 B】hdu 4932 Miaomiao's Geometry

Miaomiao's Geometry

[align=center]Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2443    Accepted Submission(s): 600
[/align]

[align=left]Problem Description[/align]

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.
 

[align=left]Input[/align]

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.
 

[align=left]Output[/align]

For each test cases , output a real number shows the answser. Please output three digit after the decimal point.
 

[align=left]Sample Input[/align]

3
3
1 2 3
3
1 2 4
4
1 9 100 10

 

[align=left]Sample Output[/align]

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.

 

题意：给定x轴上一些点（不重复），现在要选一个线段，使得能放进这些区间中，保证线段不跨过点（即线段上只能是最左边或最右边是点），并且没有线段相交，求能放进去的最大线段

思路：推理一下，只有两点之间的线段，还有线段的一半可能符合题意，然后对于每种线段，去判断一下能不能成功放进去，这步用贪心，优先放左边，不行再放右边

#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])
{
break;
}

if (c[j] - tmp >= c[j - 1])
{
if (flag[j - 1] == 2)
{
if (c[j] - c[j - 1] == tmp)
{
flag[j] = 1;
}
else if (c[j] - c[j - 1] >= 2 * tmp)
{
flag[j] = 1;
}
else if (c[j] + tmp <= c[j + 1])
{
flag[j] = 2;
}
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;
}