HDU 3934 & poj 2079 (convex hull + rotating jamming case + finding the maximum Triangle Area)

Link: http://poj.org/problem? Id = 2079 triangle

Time limit:3000 Ms		Memory limit:30000 K
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 shoshould 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

33 42 62 752 63 92 08 06 5-1

Sample output

0.5027.00

Source

Shanghai 2004 preliminary variable | | | | convex Hull, I have referenced n convex hull construction, but the construction methods are different. Some convex hull are also scanned over and over again. I don't know the difference between them and the cross multiplication, if you don't have a deep understanding of the Cross-ride, you have to re-view the rotary jamming case.

  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\n",ans);112     }113     return 0;114 }115    

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