HDU 4932 Miaomiao's Geometry(BestCoder Round #4)
2015-10-21 16:54
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[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
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:有一些点,现在需要一些相同长度的片段遮住这些点(每个点在片段的两端才算遮住),这些片段不能有重叠部分,求出满足这样的片段的最大长度。
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
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:有一些点,现在需要一些相同长度的片段遮住这些点(每个点在片段的两端才算遮住),这些片段不能有重叠部分,求出满足这样的片段的最大长度。
#include<stdio.h> #include<string.h> #include<math.h> #include<stdlib.h> #include<queue> #include<algorithm> using namespace std; const int N=1e5+10; const int M=50000; const int INF=0x3f3f3f3f; double a[100], b[200]; ///b数组保存两点之间的长度及其一半 int n; int Judge(double d) ///判断d能不能当满足条件的片段长度 { double p = a[0]; int i; for (i = 1; i < n; i++) { if (fabs(p-a[i]) < 1e-9) continue; if (p > a[i]) return 0; else if (p+d <= a[i]) p = a[i]; else p = a[i]+d; } return 1; } int main () { double Max; int T, i, k; scanf("%d", &T); while (T--) { scanf("%d", &n); for (i = 0; i < n; i++) scanf("%lf", &a[i]); memset(b, 0, sizeof(b)); Max = -INF; k = 0; sort(a, a+n); for (i = 1; i < n; i++) { b[k++] = a[i]-a[i-1]; b[k++] = (a[i]-a[i-1])/2; } for (i = 0; i < k; i++) { if (Judge(b[i])) Max = max(Max, b[i]); } printf("%.3f\n", Max); } return 0; }
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