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  • poj 2079 Triangle(旋转卡壳)

    Time Limit: 3000MS   Memory Limit: 30000K
    Total Submissions: 7637   Accepted: 2238

    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

    3
    3 4
    2 6
    2 7
    5
    2 6
    3 9
    2 0
    8 0
    6 5
    -1

    Sample Output

    0.50
    27.00

    Source

     
     
     
     
      1 #include <stdio.h>
      2 #include <string.h>
      3 #include <algorithm>
      4 #include <iostream>
      5 #include <math.h>
      6 using namespace std;
      7 #define eps 1e-8
      8 
      9 struct Point
     10 {
     11     int x,y;
     12     Point(int _x = 0, int _y = 0)
     13     {
     14         x = _x;
     15         y = _y;
     16     }
     17     Point operator -(const Point &b)const
     18     {
     19         return Point(x - b.x, y - b.y);
     20     }
     21     int operator ^(const Point &b)const
     22     {
     23         return x*b.y - y*b.x;
     24     }
     25     int operator *(const Point &b)const
     26     {
     27         return x*b.x + y*b.y;
     28     }
     29     void input()
     30     {
     31         scanf("%d%d",&x,&y);
     32     }
     33 };
     34 int dist2(Point a,Point b)//距离的平方!!!attention!! 
     35 {
     36     return (a-b)*(a-b);
     37 }
     38 
     39 
     40 int dcmp(double a)
     41 {
     42     if(fabs(a)<eps)return 0;
     43     if(a>0)return 1;
     44     else return -1; 
     45 }
     46 
     47 const int MAXN = 50010;
     48 Point list[MAXN];
     49 int Stack[MAXN],top;
     50 bool _cmp(Point p1,Point p2)
     51 {
     52     int tmp = (p1-list[0])^(p2-list[0]);
     53     if(tmp > 0)return true;
     54     else if(tmp == 0 && dist2(p1,list[0]) <= dist2(p2,list[0]))
     55         return true;
     56     else return false;
     57 }
     58 
     59 void Graham(int n)//求凸包 
     60 {
     61     Point p0;
     62     int k = 0;
     63     p0 = list[0];
     64     for(int i = 1;i < n;i++)
     65         if(p0.y > list[i].y || (p0.y == list[i].y && p0.x > list[i].x))
     66         {
     67             p0 = list[i];
     68             k = i;
     69         }
     70     swap(list[0],list[k]);
     71     sort(list+1,list+n,_cmp);
     72     if(n == 1)
     73     {
     74         top = 1;
     75         Stack[0] = 0;
     76         return;
     77     }
     78     if(n == 2)
     79     {
     80         top = 2;
     81         Stack[0] = 0;
     82         Stack[1] = 1;
     83         return;
     84     }
     85     Stack[0] = 0;
     86     Stack[1] = 1;
     87     top = 2;
     88     for(int i = 2;i < n;i++)
     89     {
     90         while(top > 1 && ((list[Stack[top-1]]-list[Stack[top-2]])^(list[i]-list[Stack[top-2]])) <= 0 )
     91             top--;
     92         Stack[top++] = i;
     93     }
     94 }
     95 
     96 double Area(Point *p,int n)
     97 {
     98     double area=0;
     99     for(int i=0; i<n; i++){
    100         area+=(p[i]^p[(i+1)%n]);
    101     }
    102     return fabs((double)(area)/2);
    103 }
    104 
    105 double rotating_calipers(Point p[],int n)
    106 {
    107     double ans = 0;
    108     Point v;
    109     int cur = 1;
    110     for(int i = 0;i < n;i++)//先转k再转j再转i!!! 
    111     {
    112         v = p[i]-p[(i+1)%n];
    113         int k,j;
    114         j=(i+1)%n;
    115         k=(j+1)%n;
    116         while(k!=i&&j!=i)
    117         {      
    118             v = p[i]-p[j];//wa无数次!忘记了j也在转啊!!!    
    119                while(dcmp(v^(p[(k+1)%n]-p[k])) < 0)    
    120             {   
    121                 k = (k+1)%n;            
    122             }
    123             Point aa[4];
    124             aa[0]=p[i];aa[1]=p[j%n];aa[2]=p[k%n];
    125             double tmp=Area(aa,3);  
    126             ans=max(ans,tmp);
    127             j=(j+1)%n;
    128         }
    129            Point aa[4];
    130         aa[0]=p[i];aa[1]=p[j%n];aa[2]=p[k%n];
    131         double tmp=Area(aa,3);  
    132         ans=max(ans,tmp);
    133     }
    134     return ans;
    135 }
    136 Point p[MAXN];
    137  
    138 int main()
    139 {
    140     int n;
    141     while(scanf("%d",&n) == 1&&n!=-1)
    142     {
    143         for(int i = 0;i < n;i++)
    144             list[i].input();
    145         Graham(n);
    146         for(int i = 0;i < top;i++)
    147         {
    148             p[i] = list[Stack[i]];
    149         }
    150         printf("%.2f
    ",rotating_calipers(p,top)) ;
    151     }
    152     return 0;
    153 }
    View Code
     
     
     
     
     
     
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  • 原文地址:https://www.cnblogs.com/skykill/p/3240481.html
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