POJ 2079 Triangle（旋转卡壳）

比眉伴天荒 2022-07-24 11:18 203阅读 0赞

POJ 2079 Triangle（旋转卡壳）：[http://poj.org/problem?id=2079][http_poj.org_problem_id_2079]

题面描述：

Triangle

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;3000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;30000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;9131</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;2704</td> 
  </tr> 
 </tbody> 
</table>

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 −10 4 <= xi, yi <= 10 4 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

题目大意：

给定一个平面上的n个点，找出以这些点为顶点的三角形且面积最大的一个，并输出它的面积。

算法分析：

由于最大的三角形总会由凸包的顶点所围成，故可以先做一个凸包，但是如果直接枚举凸包的顶点，则计算效率较低，会超时，所以此处采用旋转卡壳的算法来求解此题。

首先要枚举三角形的第一个顶点i，然后初始化第二个顶点j=i+1和第三个顶点k=j+1，循环执行k+1直到Area(i,j,k)>Area(i,j,k+1)，同时更新最大面积，旋转j，k两个点，如果Area(i,j,k)<Area(i,j,k+1)且k不等于i，则执行k=k+1，否则更新面积，并执行j=j+1，如果j=i，则跳出循环，这样旋转一周，即可求得以i为顶点的最大三角形面积，时间复杂度为O(n2).

代码实现：

#include <iostream>
    #include <algorithm>
    #include <cstdio>
    
    #define N 50005
    #define max(a,b) a>b?a:b
    
    using namespace std;
    
    struct node
    {
        int x,y;
    }dd[N];
    
    int n,stak[N],top,top1;
    bool cmp(node a,node b)
    {
        return a.x<b.x||(a.x==b.x && a.y<b.y);
    }
    bool judege_right(int o,int a,int b)
    {
        int ax=dd[a].x-dd[o].x;
        int bx=dd[b].x-dd[o].x;
        int ay=dd[a].y-dd[o].y;
        int by=dd[b].y-dd[o].y;
        return bx*ay>ax*by;
    }
    
    double area(int o,int a,int b)
    {
        int ax=dd[a].x-dd[o].x;
        int bx=dd[b].x-dd[o].x;
        int ay=dd[a].y-dd[o].y;
        int by=dd[b].y-dd[o].y;
        return abs(bx*ay-ax*by)*1.0/2.0;
    }
    
    void build_map()
    {
        int i;
        top=0;
        sort(dd,dd+n,cmp);
        stak[top++]=0;
        stak[top++]=1;
        for(i=2;i<n;i++)
        {
            stak[top++]=i;
            while(top>=3)
            {
                if(judege_right(stak[top-3],stak[top-2],stak[top-1]))
                    break;
                stak[top-2]=stak[top-1];
                top--;
            }
        }
        top1=top;
        stak[top++]=n-2;
        for(i=n-3;i>=0;i--)
        {
            stak[top++]=i;
            while(top-top1>=2)
            {
                if(judege_right(stak[top-3],stak[top-2],stak[top-1]))
                    break;
                stak[top-2]=stak[top-1];
                top--;
            }
        }
        top--;
    }
    int main()
    {
        int i,j,k;
        while(scanf("%d",&n)!=EOF)
        {
            if(n==-1)
            {
                break;
            }
            for(i=0;i<n;i++)
            {
                scanf("%d%d",&dd[i].x,&dd[i].y);
            }
            build_map();
            double ans=0;
            for(i=0;i<top;i++)
            {
                j=(i+1)%top;
                k=(j+1)%top;
                while(k!=i&&area(stak[i],stak[j],stak[k])<area(stak[i],stak[j],stak[(k+1)%top]))
                {
                    k=(k+1)%top;
                }
                if(k==i)
                    continue;
                int kk=(k+1)%top;
                while(j!=kk&&k!=i)
                {
                    ans=max(ans,area(stak[i],stak[j],stak[k]));
                    while(k!=i&&area(stak[i],stak[j],stak[k])<area(stak[i],stak[j],stak[(k+1)%top]))
                    {
                        k=(k+1)%top;
                    }
                    j=(j+1)%top;
                }
            }
            printf("%.2lf\n",ans);
        }
        return 0;
    }

[http_poj.org_problem_id_2079]: http://poj.org/problem?id=2079