POJ 2079 Triangle（旋转卡壳计算平面点集最大三角形面积）

Triangle

Time Limit: 3000MS
Memory Limit: 30000KTotal Submissions: 9673
Accepted: 2896

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

33 42 62 752 63 92 08 06 5-1

Sample Output

0.5027.00

Source

Shanghai 2004 Preliminary

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 = 50005;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;}Point p[MAXN];int main(){    int n;    while(scanf("%d",&n)==1)    {    if(n == -1)break;        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;}

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 = 50005;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[MAXN];int main(){    int n;    while(scanf("%d",&n)==1)    {    if(n == -1)break;        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://poj.org/problem?id=2079