poj 1408

题目http://poj.org/problem?id=1408

 题意：在x轴，x = 1，y轴，y = 1.这四条线段上给出一些点，然后把x轴上的点和x = 1 上的点一一对应连接，把y轴上的点和y = 1上的点也一一对应连接，在这些连接线组成的四边形（除了定点外，不含其它的点）中，找出一个面积最大的

思路：求出所有的交点，然后求每一个四边形面积，找出最大输出即可，纠结了很久都不出结果，就是因为一个点保存的时候x写成了y，真杯具。

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <iostream>
#define N 2000

using namespace std;

struct node
{
    double x;
    double y;
};
int n;
node sx[N][2];
node zy[N][2];
node point[N][N];
void init()
{
    int i;
    point[0][0].x = point[0][0].y = point[n + 1][0].y = point[0][n + 1].x = 0;
    point[n + 1][0].x = point[0][n + 1].y = point[n + 1][n + 1].x = point[n + 1][n + 1].y = 1; // ******
    for(i = 1;i <= n; i++)
    {
        scanf("%lf",&sx[i][0].x);
        sx[i][0].y = 0;
    }
    for(i = 1;i <= n; i++)
    {
        scanf("%lf",&sx[i][1].x);
        sx[i][1].y = 1;
    }
    for(i = 1;i <= n; i++)
    {
        scanf("%lf",&zy[i][0].y);
        zy[i][0].x = 0;
    }
    for(i = 1;i <= n; i++)
    {
        scanf("%lf",&zy[i][1].y);
        zy[i][1].x = 1;
    }
    for(i = 1;i <= n; i++)
    {
        point[i][0] = sx[i][0];
        point[i][n + 1] = sx[i][1];
        point[0][i] = zy[i][0];
        point[n + 1][i] = zy[i][1];
    }
}
node jiaodian(node a1,node a2,node a3,node a4)  // 四种情况求交点
{
    node p;
    double k1,k2,b1,b2;
    if(a1.x == a2.x)
    {
        if(a3.y == a4.y)
        {
            p.x = a1.x;p.y = a3.y;
        }
        else
        {
            k1 = a4.y - a3.y;
            p.x = a1.x;
            p.y = (k1 * p.x) + a3.y;
        }
    }
    else
    {
        k1 = 1 * 1.0 / (a2.x - a1.x);
        b1 = -1 * k1 * a1.x;
        if(a3.y == a4.y)
        {
            p.y = a3.y;
            p.x = (p.y - b1) / k1;
        }
        else
        {
            k2 = a4.y - a3.y;
            b2 = a3.y;
            p.x = (b2 - b1) * 1.0 / (k1 - k2);
            p.y = k2 * p.x + b2;
        }
    }
    //cout<<"P"<<p.x<<" "<<p.y<<endl<<endl;
    return p;
}
double qmianji(node a1,node a2,node a3,node a4) // 叉积求多边形面积
{
    node temp[10];                        // 参考链接http://blog.csdn.net/hbzjklqh/article/details/6707819
    temp[0] = a1;
    temp[1] = a2;
    temp[2] = a3;
    temp[3] = a4;
    int i;
    double kemp = 0,cemp = 0;
    double s;
    for(i = 0;i < 4; i++)
    {
        kemp += temp[i].x * temp[(i + 1) % 4].y;
        cemp += temp[i].y * temp[(i + 1) % 4].x;
    }
    //cout<<kemp<<" "<<cemp<<endl;
    s = fabs(kemp - cemp) / 2;
    //cout<<s<<endl;
    return s;
}
int main()
{
    int i,j;
    double mianji[N];
    freopen("data.txt","r",stdin);
    while(cin>>n,n)
    {
        memset(mianji,0,sizeof(mianji));
        init();
        for(i = 1; i <= n; i++)
        {
            for(j = 1; j <= n; j++)
            {
                point[i][j] = jiaodian(sx[i][0],sx[i][1],zy[j][0],zy[j][1]);
            }
        }
        double maxx = 0;
        int k = 0;
        for(i = 0; i <= n; i++)
        {
            for(j = 0; j <= n; j++)
            {
                mianji[k++] = qmianji(point[i][j],point[i + 1][j],point[i + 1][j + 1],point[i][j + 1]);
            }
        }
        for(i = 0; i <= (n + 1) * (n + 1) ; i++)
        {
            if(maxx < mianji[i]) maxx = mianji[i];
        }
        printf("%.6lf\n",maxx);
    }
    return 0;
}