题目链接:http://poj.org/problem?id=2079
Description
Input
Output
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
题意:
求最大三角形面积;
PS:
代码如下:
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <math.h>
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);
}
};
int dist2(Point a,Point b)
{
return (a-b)*(a-b);
}
const int MAXN = 50010;
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[0],list[k]);
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,求三角形最大面积
int rotating_calipers(Point p[],int n)
{
int ans = 0;
Point v;
int cur = 1;
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]];
int ans = rotating_calipers(p,top);
printf("%.2lf\n",ans/2.0);
}
return 0;
}POJ 2079 Triangle(凸包_旋转卡壳之最大三角形面积)
原文:http://blog.csdn.net/u012860063/article/details/44628601