poj1408

题意：给出一个正方形，每条边上有n个点，把这些点作为端点拉线组成的网格（整个正方形由一些四边行拼成），求最大格的面积。

分析：计算几何，求出所有线的交点，分别计算每个格的面积。

在求线段交点时可以用叉积面积的方式，求ac,bd交点：

void intersection(Point &a, Point &b, Point &c, Point &d, Point &ret)
{
    double s1 = xmulti(a, c, b);
    double s2 = xmulti(a, c, d);
    ret.x = (s1 * d.x - s2 * b.x) / (s1 - s2);
    ret.y = (s1 * d.y - s2 * b.y) / (s1 - s2);
}

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
usingnamespace std;

#define maxn 50

struct Point
{
    double x, y;
} point[maxn][maxn];

double ans;
int n;

void input()
{
    n++;
    point[0][0].x =0;
    point[0][0].y =0;
    point[0][n].x =1;
    point[0][n].y =0;
    point[n][0].x =0;
    point[n][0].y =1;
    point[n][n].x =1;
    point[n][n].y =1;
    for (int i =1; i < n; i++)
    {
        scanf("%lf", &point[0][i].x);
        point[0][i].y =0;
    }
    for (int i =1; i < n; i++)
    {
        scanf("%lf", &point[n][i].x);
        point[n][i].y =1;
    }
    for (int i =1; i < n; i++)
    {
        scanf("%lf", &point[i][0].y);
        point[i][0].x =0;
    }
    for (int i =1; i < n; i++)
    {
        scanf("%lf", &point[i][n].y);
        point[i][n].x =1;
    }
}

double xmulti(Point &p0, Point &p1, Point &p2)
{
    return ((p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y));
}

void intersection(Point &a, Point &b, Point &c, Point &d, Point &ret)
{
    double s1 = xmulti(a, c, b);
    double s2 = xmulti(a, c, d);
    ret.x = (s1 * d.x - s2 * b.x) / (s1 - s2);
    ret.y = (s1 * d.y - s2 * b.y) / (s1 - s2);
}

double area(Point &a, Point &b, Point &c, Point &d)
{
    double s1 = xmulti(a, c, b);
    double s2 = xmulti(a, c, d);
    return (abs(s1) + abs(s2)) /2;
}

void work()
{
    for (int i =1; i < n; i++)
        for (int j =1; j < n; j++)
            intersection(point[i][0], point[0][j], point[i][n], point[n][j], point[i][j]);
    ans =-1;
    for (int i =0; i < n; i++)
        for (int j =0; j < n; j++)
        {
            double temp = area(point[i][j], point[i +1][j], point[i +1][j +1], point[i][j +1]);
            ans = max(ans, temp);
        }
}

int main()
{
    //freopen("t.txt", "r", stdin);
while (scanf("%d", &n), n)
    {
        input();
        work();
        printf("%.6f\n", ans);
    }
    return0;
}