POJ 2079 Triangle

 

Triangle

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 4224		Accepted: 1149

Description

Given n distinct points on a plane, your task is to find the triangle that have the maximum area, whose vertices are from the given points.

Input

The input consists of several test cases. The first line of each test case contains an integer n, indicating the number of points on the plane. Each of the following n lines contains two integer xi and yi, indicating the ith points. The last line of the input is an integer −1, indicating the end of input, which should not be processed. You may assume that 1 <= n <= 50000 and −10 ⁴ <= xi, yi <= 10 ⁴ for all i = 1 . . . n.

Output

For each test case, print a line containing the maximum area, which contains two digits after the decimal point. You may assume that there is always an answer which is greater than zero.

Sample Input

3
3 4
2 6
2 7
5
2 6
3 9
2 0
8 0
6 5
-1

Sample Output

0.50
27.00

Source

Shanghai 2004 Preliminary

/* 第一个凸包+旋转卡壳,旋转卡壳很NB啊,需要系统学习一下,名字挺炫的 首先很容易想到具有最大面积的三角形的顶点肯定是凸包上的顶点,所以需要首先求凸包,然后在求凸包的 基础上采用旋转卡壳来计算最大面积.唉,这次又把凸包的内层while循环给丢了直接贡献了5个WA,FAINTTTTT... 旋转卡壳不想多说了这里给出一些资料,以后慢慢看 (1)求凸包最大直径的旋转卡壳算法, 可以参照 http://www.cppblog.com/staryjy/archive/2009/11/19/101412.html (2)系统学习旋转卡壳可以参考 http://cgm.cs.mcgill.ca/~orm/rotcal.frame.html (3)本题的旋转卡壳算法如下: 枚举三角形的第一个顶点i, 然后初始第二个顶点j = i + 1,第三个顶点k = j + 1， 循环k + 1直到差积(j, k, i) > 差积(j, k+1, i) 接下来开始旋转j和k 如果差积(j,k,i) < 差积(j,k+1,i) 且 k != i, 然后更新k = k + 1，否则转到下一步 更新最大面积 和j = j + 1,当j = i时跳出循环 时间复杂度是O(n^2) */ #include <iostream> #include <algorithm> #include <cmath> #define MAX_N 50005 #define EPS 1e-8 using namespace std; //存储凸包点在nodes里的下标 struct p { int index; }pol[MAX_N]; int polnum; //存储输入点 struct node { double x, y; }nodes[MAX_N]; int noden, sindex; double sx, sy; //向量(n1, n3) 与向量 (n2, n3)的差积 double crossMult(const node &n1, const node &n2, const node &n3) { return (n1.x - n3.x) * (n2.y - n3.y) - (n2.x - n3.x) * (n1.y - n3.y); } //点n1与nodes[0]的距离的平方 double dist(const node &n1) { return (n1.x - nodes[0].x) * (n1.x - nodes[0].x) + (n1.y - nodes[0].y) * (n1.y - nodes[0].y); } //计算凸包之前的排序比较函数 bool compare(const node &n1, const node &n2) { double type = crossMult(n1, n2, nodes[0]); if(type < -EPS) return false; else if(type > EPS) return true; else return dist(n1) <= dist(n2); } //计算凸包, 逆时针排序, 共线点需要加上 void getPoly() { polnum = 0; pol[polnum++].index = 0; pol[polnum++].index = 1; for(int i = 2; i < noden; i++) { while(true) //唉,这次又把这儿给漏了 { int index1 = pol[--polnum].index; int index0 = pol[--polnum].index; double type = crossMult(nodes[i], nodes[index1], nodes[index0]); pol[polnum++].index = index0; if(type <= -EPS) { pol[polnum++].index = index1; break; } if(polnum == 1) break; } pol[polnum++].index = i; } } //旋转卡壳 double getMaxSize() { int i, j, k; double maxSize = 0; //遍历i for(i = 0; i < polnum; i++) { //初始化j, k分别为下一个和下下个凸包上的点 j = (i + 1) % polnum; k = (j + 1) % polnum; double c1, c2; //旋转k直到k == i或者三角形(i, j, k+1)的面积不再增加 while(k != i) { double c1 = crossMult(nodes[pol[j].index], nodes[pol[k].index], nodes[pol[i].index]); double c2 = crossMult(nodes[pol[j].index], nodes[pol[(k + 1) % polnum].index], nodes[pol[i].index]); if(c1 + EPS < c2) k = (k + 1) % polnum; else break; } if(k == i) continue; int kk = (k + 1) % polnum; //旋转j, k直到j == kk 或者k == i while(j != kk && k != i) { //更新最大面积 c1 = crossMult(nodes[pol[j].index], nodes[pol[k].index], nodes[pol[i].index]); if(c1 > maxSize + EPS) maxSize = c1; //旋转k直到 k == i 或者三角形(i, j, k+1)的面积不再增加 while(k != i) { c1 = crossMult(nodes[pol[j].index], nodes[pol[k].index], nodes[pol[i].index]); c2 = crossMult(nodes[pol[j].index], nodes[pol[(k + 1) % polnum].index], nodes[pol[i].index]); if(c1 + EPS < c2) k = (k + 1) % polnum; else break; } j = (j + 1) % polnum; } } return maxSize / 2.0; } int main() { int i; while(scanf("%d", &noden) && noden != -1) { sx = sy = INT_MAX; for(i = 0; i < noden; i++) { scanf("%lf%lf", &nodes[i].x, &nodes[i].y); if(nodes[i].y + EPS < sy || (fabs(nodes[i].y - sy) < EPS && nodes[i].x + EPS < sx)) { sx = nodes[i].x; sy = nodes[i].y; sindex = i; } } if(noden <= 2) printf("0.00/n"); else { if(sindex != 0) { int tempx = nodes[0].x, tempy = nodes[0].y; nodes[0].x = nodes[sindex].x; nodes[0].y = nodes[sindex].y; nodes[sindex].x = tempx; nodes[sindex].y = tempy; } sort(&nodes[1], &nodes[1] + noden - 1, compare); getPoly(); printf("%.2lf/n", getMaxSize()); } } return 0; }