http://poj.org/problem?id=1408
题意：给出 a1 a2 ... an
b1 b2 ... bn
c1 c2 ... cn
d1 d2 ... dn 这些点，求这些对应点连线形成的小四边形的最大面积。
思路：将所有的交点求出，同已知点一起存入二维矩阵中，枚举每个小四边形，求出其面积，找出最大的即可。
1 #include <stdio.h> 2 #include <stdlib.h> 3 #include <math.h> 4 const double eps=1e-8;//设置精度 5 #define max1(a1,b1) (double)a1-(double)b1>eps?(double)a1:(double)b1 6 struct point//点 7 { 8 double x; 9 double y; 10 } map[120][120];//用二维数组的结构体存储所有的点 11 struct line//线 12 { 13 double a,b,c; 14 }; 15 line getline(point p1,point p2)//由两点求直线ax+by+c=0 16 { 17 line tmp; 18 tmp.a = p1.y-p2.y; 19 tmp.b = p2.x-p1.x; 20 tmp.c = p1.x*p2.y-p2.x*p1.y; 21 return tmp; 22 } 23 point getIntersect(line L1,line L2)//求两直线交点 24 { 25 point tmp; 26 tmp.x = (L1.b*L2.c-L2.b*L1.c)/(L1.a*L2.b-L2.a*L1.b); 27 tmp.y = (L1.c*L2.a-L2.c*L1.a)/(L1.a*L2.b-L2.a*L1.b); 28 return tmp; 29 } 30 double area_polygon(int n,point* p)//求多边形面积 31 { 32 double s1=0,s2=0; 33 int i; 34 for (i=0; i<n; i++) 35 { 36 s1+=p[(i+1)%n].y*p[i].x; 37 s2+=p[(i+1)%n].y*p[(i+2)%n].x; 38 } 39 return fabs(s1-s2)/2; 40 } 41 int main() 42 { 43 int n; 44 while(~scanf("%d",&n)&&n) 45 { 46 double Max = eps; 47 double a,b,c,d; 48 map[0][0].x = 0;//初始化已知的四个顶点的坐标 49 map[0][0].y = 0; 50 map[0][n+1].x = 1; 51 map[0][n+1].y = 0; 52 map[n+1][0].x = 0; 53 map[n+1][0].y = 1; 54 map[n+1][n+1].x = 1; 55 map[n+1][n+1].y = 1; 56 for (int j = 1; j <= n; j++)//输入a1 a2 ... an，并存储其坐标 57 { 58 scanf("%lf",&a); 59 map[0][j].x = a; 60 map[0][j].y = 0; 61 } 62 for (int j = 1; j <= n; j++)//输入b1 b2 ... bn，并存储其坐标 63 { 64 scanf("%lf",&b); 65 map[n+1][j].x = b; 66 map[n+1][j].y = 1; 67 } 68 for (int i = 1; i <= n; i++)//输入c1 c2 ... cn，并存储其坐标 69 { 70 scanf("%lf",&c); 71 map[i][0].x = 0; 72 map[i][0].y = c; 73 } 74 for (int i =1; i <= n; i++)//输入d1 d2 ... dn，并存储其坐标 75 { 76 scanf("%lf",&d); 77 map[i][n+1].x = 1; 78 map[i][n+1].y = d; 79 } 80 for (int i = 1; i <= n; i++) 81 for (int j = 1; j <= n; j++) 82 { 83 line L1 = getline(map[i][0],map[i][n+1]);//枚举所有可相交的直线 84 line L2 = getline(map[0][j],map[n+1][j]); 85 point tmp = getIntersect(L1,L2);//求这两条线的交点 86 map[i][j].x = tmp.x;//将求得的交点坐标存入相应的数组中 87 map[i][j].y = tmp.y; 88 89 } 90 point p[5];//存储四边形的顶点 91 for (int i = 1; i <= n+1; i++) 92 { 93 94 for (int j = 1; j <= n+1; j++)//枚举所有四边形右上角的顶点 95 { 96 p[0] = map[i][j];//以map[i][j]为右上角的顶点组成的四边形的各点 97 p[1] = map[i][j-1]; 98 p[2] = map[i-1][j-1]; 99 p[3] = map[i-1][j]; 100 double s = area_polygon(4,p);//求四边形的面积 101 Max = max1(s,Max);//求最大面积 102 } 103 104 } 105 printf("%.6f ",Max); 106 } 107 return 0; 108 }