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
題意：這道題是給出一個圓，上面有n個點，從中選出m個點，選擇一種方案使得構成的m多邊形面積最大（m個點依次順序連接形成多邊形）。
題解：（動態規劃很多時候都需要枚舉子問題的狀態，從而推出所需問題的最優策略，這樣，對於子問題的枚舉策略就顯得至關重要。既然是枚舉，所有的情況都不能遺漏，枚舉條件的選取恰當與否決定了問題能否順利解決）這題的數據規模是40，因爲涉及到最大面積，考慮動態規劃，根據題意，猜測是區間DP，對於DP參數的設定：
dp(i, j, k) 表示以i開始，以j結束的擁有k個點的多邊形的最大面積，這昂最總答案是max{dp(i, j, m)}.
狀態轉移方程：dp(i, j, k) = max{ dp(i, j, k) , dp(i, q, k - 1) + Triangle(i, q, j) }.
代碼：
#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 50
int 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;
}
