hdu 3934&&poj 2079 （凸包+旋转卡壳+求最大三角形面积）

链接：http://poj.org/problem?id=2079

Triangle

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 8173		Accepted: 2423

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 −10⁴<= xi, yi <= 10⁴ 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

Sample Output

0.50
27.00

Source

Shanghai 2004 Preliminary

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乱起八糟的凸包，参考了n个凸包构建，但是构建方式都不同，有的凸包都还倒过来扫一遍，，，，，，不知道其中的差别

还有叉乘，叉乘没有深入理解

旋转卡壳还要重新看

  1 #include <math.h>
  2 #include <stdio.h>
  3 #include <string.h>
  4 #include <stdlib.h>
  5 #include <iostream>
  6 #include <algorithm>
  7 
  8 using namespace std;
  9 
 10 #define eps 1e-8
 11 #define MAXX 1000010
 12 
 13 typedef struct point
 14 {
 15 
 16     double x;
 17     double y;
 18 }point;
 19 
 20 bool dy(double x,double y){
 21     return x>y+eps; }
 22 bool xy(double x,double y){
 23     return x<y-eps; }
 24 bool dyd(double x,double y){
 25     return x>y-eps; }
 26 bool xyd(double x,double y){
 27     return x<y+eps; }
 28 bool dd(double x,double y){
 29     return fabs(x-y)<eps; }
 30 
 31 double crossProduct(point a,point b,point c)
 32 {
 33     return (c.x-a.x)*(b.y-a.y)-(c.y-a.y)*(b.x-a.x);
 34 }
 35 
 36 double dist(point a,point b)
 37 {
 38 
 39     return sqrt((b.x-a.x)*(b.x-a.x)+(b.y-a.y)*(b.y-a.y));
 40 }
 41 
 42 point c[MAXX];
 43 point stk[MAXX];
 44 int top;
 45 
 46 bool cmp(point a,point b)
 47 {
 48 
 49     double len=crossProduct(c[0],a,b);
 50     if(dd(len,0.0))
 51         return xy(dist(c[0],a),dist(c[0],b));
 52     return xy(len,0.0);
 53 }
 54 
 55 double max(double x,double y)
 56 {
 57 
 58     return xy(x,y) ? y : x;
 59 }
 60 
 61 void Graham(int n)
 62 {
 63 
 64     int tmp=0;
 65     for(int i=1; i<n; i++)
 66     {
 67 
 68         if(xy(c[i].x,c[tmp].x) || dd(c[i].x,c[tmp].x) && xy(c[i].y,c[tmp].y))
 69             tmp=i;
 70     }
 71     swap(c[0],c[tmp]);
 72     sort(c+1,c+n,cmp);
 73     stk[0]=c[0];
 74     stk[1]=c[1];
 75     top=1;
 76     for(int i=2; i<n; i++)
 77     {
 78         while(top>=1 && xyd(crossProduct(stk[top],stk[top-1],c[i]),0.0))
 79             top--;
 80         stk[++top]=c[i];
 81     }
 82 }
 83 
 84 double rotating(int n)
 85 {
 86     int j=1,k=2;
 87     double ans=0.0;
 88     stk[n]=stk[0];
 89     for(int i=0; i<n; i++)
 90     {
 91         while(dy(fabs(crossProduct(stk[(k+1)%n],stk[i],stk[j])),fabs(crossProduct(stk[k],stk[i],stk[j]))))
 92             k=(k+1)%n;
 93         while(dy(fabs(crossProduct(stk[k],stk[i],stk[(j+1)%n])),fabs(crossProduct(stk[k],stk[i],stk[j]))))
 94             j=(j+1)%n;
 95         ans=max(ans,fabs(crossProduct(stk[k],stk[i],stk[j])));
 96     }
 97     return ans*0.5;
 98 }
 99 
100 int main()
101 {
102 
103     int i,j,n;
104     while(scanf("%d",&n)!=EOF&&n != -1)
105     {
106 
107         for(i=0; i<n; i++)
108             scanf("%lf%lf",&c[i].x,&c[i].y);
109         Graham(n);//printf("%d**",top);
110         double ans=rotating(top+1);
111         printf("%.2lf
",ans);
112     }
113     return 0;
114 }
115

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