HDU 3934 Summer holiday(转卡壳计算平面点集最大三角形面积)

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#旋转卡壳 #多边形面积 #凸包 #三角形

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Summer holiday

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1923 Accepted Submission(s): 694

Problem Description

Summer holiday was coming! Xiaomao went back to his hometown where he yearn day and night, his hometown has picturesque scenery. There is a big forest beside his village. There are n trees in the forest.
Now they want to across the forest with a rope (the rope won't cross). Try to find 3 trees in this tree on the rope which can make the area of the surrounded largest. Work out the area of it.

Input

The input will consist of several test cases. The first line contains a positive integer N（3<=N<=10^6）, the number of trees, followed N lines, each gives the (xi, yi ) coordinates.

Output

Print the largest area, one number a line with two decimal places.

Sample Input

Sample Output

  
   0.50

Source

2011 Multi-University Training Contest 10 - Host by HRBEU

题目大意：给定N个点，求这些点中任意三点围成的最大三角形的面积。

解题思路：先求出一个凸包包围这些点，然后用旋转卡壳求凸包内任意三角形面积比大小。

凸包模版

const int MAXN = 1000005;
Point list[MAXN];
int Stack[MAXN],top;
bool _cmp(Point p1,Point p2)
{
	int tmp = (p1-list[0])^(p2-list[0]);
    if(tmp > 0)return true;
    else if(tmp == 0 && dist2(p1,list[0]) <= dist2(p2,list[0]))
    return true;
    else return false;
}
void Graham(int n)
{
    Point p0;
    int k = 0;
    p0 = list[0];
    for(int i = 1;i < n;i++)
    if(p0.y > list[i].y || (p0.y == list[i].y && p0.x > list[i].x))
    {
        p0 = list[i];
        k = i;
    }
    swap(list[k],list[0]);
    sort(list+1,list+n,_cmp);
    if(n == 1)
    {
        top = 1;
        Stack[0] = 0;
        return;
    }
    if(n == 2)
    {
        top = 2;
        Stack[0] = 0; Stack[1] = 1;
        return;
    }
    Stack[0] = 0; Stack[1] = 1;
    top = 2;
    for(int i = 2;i < n;i++)
    {
        while(top > 1 &&((list[Stack[top-1]]-list[Stack[top-2]])^(list[i]-list[Stack[top-2]])) <= 0)
        top--;
        Stack[top++] = i;
    }
}

旋转卡壳计算平面点集最大三角形面积模版

int rotating_calipers(Point p[],int n)
{
    int ans = 0;
    Point v;
    for(int i = 0;i < n;i++)
    {
        int j = (i+1)%n;
        int k = (j+1)%n;
        while(j != i && k != i)
        {
            ans = max(ans,abs((p[j]-p[i])^(p[k]-p[i])));
            while( ((p[i]-p[j])^(p[(k+1)%n]-p[k])) < 0 )
            k = (k+1)%n;
            j = (j+1)%n;
        }
    }
    return ans;
}

以下为AC代码

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm> 
using namespace std;
struct Point
{
    int x,y;
    Point(int _x = 0,int _y = 0)
    {
        x = _x; y = _y;
    }
    Point operator -(const Point &b)const
    {
        return Point(x - b.x, y - b.y);
    }
    int operator ^(const Point &b)const
    {
        return x*b.y - y*b.x;
    }
    int operator *(const Point &b)const
    {
        return x*b.x + y*b.y;
    }
    void input()
    {
        scanf("%d%d",&x,&y);
    }
};
//距离的平方
int dist2(Point a,Point b)
{
    return (a-b)*(a-b);
}
const int MAXN = 1000005;
Point list[MAXN];
int Stack[MAXN],top;
bool _cmp(Point p1,Point p2)
{
	int tmp = (p1-list[0])^(p2-list[0]);
    if(tmp > 0)return true;
    else if(tmp == 0 && dist2(p1,list[0]) <= dist2(p2,list[0]))
    return true;
    else return false;
}
void Graham(int n)
{
    Point p0;
    int k = 0;
    p0 = list[0];
    for(int i = 1;i < n;i++)
    if(p0.y > list[i].y || (p0.y == list[i].y && p0.x > list[i].x))
    {
        p0 = list[i];
        k = i;
    }
    swap(list[k],list[0]);
    sort(list+1,list+n,_cmp);
    if(n == 1)
    {
        top = 1;
        Stack[0] = 0;
        return;
    }
    if(n == 2)
    {
        top = 2;
        Stack[0] = 0; Stack[1] = 1;
        return;
    }
    Stack[0] = 0; Stack[1] = 1;
    top = 2;
    for(int i = 2;i < n;i++)
    {
        while(top > 1 &&((list[Stack[top-1]]-list[Stack[top-2]])^(list[i]-list[Stack[top-2]])) <= 0)
        top--;
        Stack[top++] = i;
    }
}
//旋转卡壳计算平面点集最大三角形面积
int rotating_calipers(Point p[],int n)
{
    int ans = 0;
    Point v;
    for(int i = 0;i < n;i++)
    {
        int j = (i+1)%n;
        int k = (j+1)%n;
        while(j != i && k != i)
        {
            ans = max(ans,abs((p[j]-p[i])^(p[k]-p[i])));
            while( ((p[i]-p[j])^(p[(k+1)%n]-p[k])) < 0 )
            k = (k+1)%n;
            j = (j+1)%n;
        }
    }
    return ans;
}
Point p[1000005];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i = 0;i < n;i++)list[i].input();
        Graham(n);
        for(int i = 0;i < top;i++)p[i] = list[Stack[i]];
        printf("%.2f\n",(double)rotating_calipers(p,top)/2);
    }
    return 0;
}

原题链接： http://acm.hdu.edu.cn/showproblem.php?pid=3934