ZOJ 1679

Telescope

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Time Limit: 5 Seconds      Memory Limit: 32768 KB

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Dr. Extreme experimentally made an extremely precise telescope to investigate extremely curious phenomena at an extremely distant place. In order to make the telescope so precise as to investigate phenomena at such an extremely distant place, even quite a small distortion is not allowed. However, he forgot the influence of the internal gas affected by low-frequency vibration of magnetic flux passing through the telescope. The cylinder of the telescope is not affected by the low-frequency vibration, but the internal gas is.

The cross section of the telescope forms a perfect circle. If he forms a coil by putting extremely thin wire along the (inner) circumference, he can measure (the average vertical component of) the temporal variation of magnetic flux: such measurement would be useful to estimate the influence. But points on the circumference at which the wire can be fixed are limited; furthermore, the number of special clips to fix the wire is also limited. To obtain the highest sensitivity, he wishes to form a coil of a polygon shape with the largest area by stringing the wire among carefully selected points on the circumference.

Your job is to write a program which reports the maximum area of all possible m-polygons (polygons with exactly m vertices) each of whose vertices is one of the n points given on a circumference with a radius of 1. An example of the case n = 4 and m = 3 is illustrated below.

In the figure above, the equations such as "p1 = 0.0" indicate the locations of the n given points, and the decimals such as "1.000000" on m-polygons indicate the areas of m-polygons.

Parameter pi denotes the location of the i-th given point on the circumference (1 <= i <= n). The location p of a point on the circumference is in the range 0 <= p < 1, corresponding to the range of rotation angles from 0 to 2*PI radians. That is, the rotation angle of a point at p to the point at 0 equals 2*PI*p radians. (PI is the circular constant 3.14159265358979323846....)

You may rely on the fact that the area of an isosceles triangle ABC (AB = AC = 1) with an interior angle BAC of �� radians (0 < a< PI) is sin(a)/2, and the area of a polygon inside a circle with a radius of 1 is less than PI.

Input

The input consists of multiple subproblems followed by a line containing two zeros that indicates the end of the input. Each subproblem is given in the following format.

n m
p1 p2 ... pn

n is the number of points on the circumference (3 <= n <= 40). m is the number of vertices to form m-polygons (3 <= m <= n). The locations of n points, p1, p2, ..., pn, are given as decimals and they are separated by either a space character or a newline. In addition, you may assume that 0 <= p1 < p2 < ... < pn < 1.

Output

For each subproblem, the maximum area should be output, each in a separate line. Each value in the output may not have an error greater than 0.000001 and its fractional part should be represented by 6 decimal digits after the decimal point.

Sample Input

4 3
0.0 0.25 0.5 0.666666666666666666667
4 4
0.0 0.25 0.5 0.75
30 15
0.00 0.03 0.06 0.09 0.12 0.15 0.18 0.21 0.24 0.27
0.30 0.33 0.36 0.39 0.42 0.45 0.48 0.51 0.54 0.57
0.61 0.64 0.66 0.69 0.72 0.75 0.78 0.81 0.84 0.87
40 20
0.351 0.353 0.355 0.357 0.359 0.361 0.363 0.365 0.367 0.369
0.371 0.373 0.375 0.377 0.379 0.381 0.383 0.385 0.387 0.389
0.611 0.613 0.615 0.617 0.619 0.621 0.623 0.625 0.627 0.629
0.631 0.633 0.635 0.637 0.639 0.641 0.643 0.645 0.647 0.649
0 0

Sample Output

1.183013
2.000000
3.026998
0.253581

  题意：这道题是给出一个圆，上面有n个点，从中选出m个点，选择一种方案使得构成的m多边形面积最大（m个点依次顺序连接形成多边形）。

题解：（动态规划很多时候都需要枚举子问题的状态，从而推出所需问题的最优策略，这样，对于子问题的枚举策略就显得至关重要。既然是枚举，所有的情况都不能遗漏，枚举条件的选取恰当与否决定了问题能否顺利解决）这题的数据规模是40，因为涉及到最大面积，考虑动态规划，根据题意，猜测是区间DP，对于DP参数的设定：

dp(i, j, k) 表示以i开始，以j结束的拥有k个点的多边形的最大面积，这昂最总答案是max{dp(i, j, m)}.

状态转移方程：dp(i, j, k) = max{ dp(i, j, k) , dp(i, q, k - 1) + Triangle(i, q, j) }.

代码：

#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const double PI = acos(-1.0);const int INF = 0x3f3f3f3f;#define N 50int n, m;double dp[N][N][N], _area[N][N][N];struct Point{double x, y;}P[N];double cross(Point a, Point b, Point c)  {      return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);  }  double area(Point a, Point b, Point c)  {      return fabs(cross(a, b, c)/2.0);  } void init(){for(int i = 0; i < n; i++)for(int j = 0; j < n; j++)for(int k = 0; k < n; k++){_area[i][j][k] = area(P[i], P[j], P[k]);}memset(dp, 0, sizeof dp);}void DP(){for(int i = 0; i < n; i++)for(int j = i + 2; j < n; j++)for(int k = 3; k <= m; k++)for(int q = i + k - 2 ; q < j; q++)dp[i][j][k] = max(dp[i][j][k], dp[i][q][k-1] + _area[i][j][q]);double ans = -INF;for(int i = 0; i < n; i++)for(int j = i + m - 1; j < n; j++){ans = max(ans, dp[i][j][m]);}printf("%.6lf\n", ans);}int main(){// freopen("2016.txt", "r", stdin);while(scanf("%d%d", &n, &m) && n + m){for(int i = 0; i < n; i++){double tmp;scanf("%lf", &tmp);tmp = 2 * PI * tmp;P[i].x = cos(tmp);P[i].y = sin(tmp);}init();DP();}return 0;}