Sicily 1829/1948. Largest Area

1829. Largest Area

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

There are N (0 < N < 30) points on a plane. Please select m (0 < m < K < 10) of them

P₁, P₂… P_m

so that they make up a polygon with largest area. These m points are all vertexes of the polygon.

Input

The first line contains an integer indicating the number of test cases.

       For each test case, there are two integers N and K in the first line. The following N lines contain a pair of integers X and Y (-1000 < X, Y < 1000) per line. Each pair indicates the coordinates of a point. There are no two points with the same coordinates.

       NO bland line between two test cases.

Output

For each test case, print the largest area that can be achieved. The result should be rounded to two digits after decimal. There is NO bland line between two test cases.

Sample Input

16 3-1 0 0 01 0-1 10 11 1

Sample Output

1.00

// Problem#: 1829// Submission#: 3589892// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University#include <stdio.h>#include <math.h>#include <stdlib.h>#include <string.h>#include <algorithm>using namespace std;const int MAXN = 300;const double eps = 1e-4;struct node {    double x, y;    double dist, angle;}p[MAXN], ch[MAXN * 2];int id[MAXN];int stack[MAXN], top;int n, k, chn;double C(node & p1, node & p2, node & p0) {    return (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y);}double area(node & p1, node & p2, node p3) {    return fabs((p1.x * p2.y - p1.y * p2.x) + (p2.x * p3.y - p2.y * p3.x) + (p3.x * p1.y - p3.y * p1.x)) / 2.0;}void init() {    int i, mid;    mid = 0;    for (i = 0; i < n; i++) {        id[i] = i;        if (p[i].y - p[mid].y < eps)            if (fabs(p[i].y - p[mid].y) < eps) {                if (p[i].x - p[mid].x < eps) mid = i;            } else mid = i;    }    id[0] = mid;    id[mid] = 0;    p[id[0]].dist = 0;    p[id[0]].angle = 0;    for (i = 1; i < n; i++) {        p[id[i]].dist = sqrt((p[id[i]].x - p[id[0]].x) * (p[id[i]].x - p[id[0]].x) + (p[id[i]].y - p[id[0]].y) * (p[id[i]].y - p[id[0]].y));        p[id[i]].angle = acos((p[id[i]].x - p[id[0]].x) / p[id[i]].dist);    }}bool cmp(const int & i, const int & j) {    if (p[i].angle - p[j].angle < eps)        if (fabs(p[i].angle - p[j].angle) < eps)            if (p[i].dist - p[j].dist < eps)                if (fabs(p[i].dist - p[j].dist) < eps)                    return true;                else                    return true;            else                 return false;        else             return true;    else         return false;}void graham() {    int i, t;    top = 0;    stack[top] = id[0];    if (n == 1) return;    stack[++top] = id[1];    if (n == 2) return;    stack[++top] = id[2];    if (n == 3) return;    for (i = 3; i < n; i++) {        while (C(p[id[i]], p[stack[top]], p[stack[top - 1]]) > eps) top--;        stack[++top] = id[i];    }    t = stack[top];    for (i = n - 2; i >= 3; i--) {        if (fabs(p[t].angle) < eps) break;        if (fabs(p[id[i]].angle - p[t].angle) < eps) stack[++top] = id[i];    }}void convexHull() {    init();    sort(id, id + n, cmp);    graham();    for (int i = 0; i <= top; i++) ch[i] = p[stack[i]];    chn = top + 1;}double maxArea[MAXN][MAXN];double maxResult;double dynamic() {    for (int i = 0; i < chn; i++) ch[i + chn] = ch[i];    maxResult = 0;    for (int i = 0; i < chn; i++) {        memset(maxArea, 0, sizeof(maxArea));        for (int j = i + 1; j < i + chn; j++) {            maxArea[j][0] = -1;            maxArea[j][1] = 0;            for (int t = 2; t <= min(j - i + 1, k - 1); t++) {                maxArea[j][t] = 0;                for (int q = i + t - 1; q <= j - 1; q++)                    if (maxArea[q][t - 1] != -1)                        maxArea[j][t] = max(maxArea[q][t - 1] + area(ch[i], ch[j], ch[q]), maxArea[j][t]);                if (maxArea[j][t] > maxResult) maxResult = maxArea[j][t];            }        }    }    return maxResult;}int main() {    int testcase;    scanf("%d", &testcase);    while (testcase--) {        scanf("%d%d", &n, &k);        for (int i = 0; i < n; i++)            scanf("%lf%lf", &p[i].x, &p[i].y);        convexHull();        printf("%.2lf\n", dynamic());    }    return 0;}