hdu 5251 矩形面积（百度之星初赛1）（计算几何）

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

const int MAXN=5000;
const double PI=acos(-1.0);

struct point
{
int x,y;
};
point list[MAXN];
int stack[MAXN],top;

int cross(point p0,point p1,point p2) //计算叉积  p0p1 X p0p2
{
return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x);
}
double dis(point p1,point p2)  //计算 p1p2的 距离
{
return sqrt((double)(p2.x-p1.x)*(p2.x-p1.x)+(p2.y-p1.y)*(p2.y-p1.y));
}
bool cmp(point p1,point p2) //极角排序函数 ， 角度相同则距离小的在前面
{
int tmp=cross(list[0],p1,p2);
if(tmp>0) return true;
else if(tmp==0&&dis(list[0],p1)<dis(list[0],p2)) return true;
else return false;
}
void init(int n) //输入，并把  最左下方的点放在 list[0]  。并且进行极角排序
{
int i,k;
point p0;
scanf("%d%d",&list[0].x,&list[0].y);
p0.x=list[0].x;
p0.y=list[0].y;
k=0;
for(i=1;i<n;i++)
{
scanf("%d%d",&list[i].x,&list[i].y);
if( (p0.y>list[i].y) || ((p0.y==list[i].y)&&(p0.x>list[i].x)) )
{
p0.x=list[i].x;
p0.y=list[i].y;
k=i;
}
}
list[k]=list[0];
list[0]=p0;

sort(list+1,list+n,cmp);
}

void graham(int n)
{
int i;
if(n==1) {top=0;stack[0]=0;}
if(n==2)
{
top=1;
stack[0]=0;
stack[1]=1;
}
if(n>2)
{
for(i=0;i<=1;i++) stack[i]=i;
top=1;

for(i=2;i<n;i++)
{
while(top>0&&cross(list[stack[top-1]],list[stack[top]],list[i])<=0) top--;
top++;
stack[top]=i;
}
}
}

int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T;
int iCase = 0;
scanf("%d",&T);
int n;
while(T--){
iCase++;
scanf("%d",&n);
init(4*n);
graham(4*n);
double ans = 1e20;
stack[top+1] = 0;
for(int i = 0;i <= top;i++){
double x1 = list[stack[i]].x;
double y1 = list[stack[i]].y;
double x2 = list[stack[i+1]].x;
double y2 = list[stack[i+1]].y;
double Min1 = 1e30, Max1 = -1e30;
double Min2 = 1e30, Max2 = -1e30;
for(int i = 0;i <= top;i++){
double x0 = list[stack[i]].x;
double y0 = list[stack[i]].y;
double tmp1 = -(y2-y1)*x0 + (x2-x1)*y0;
double tmp2 = -(x2-x1)*x0 - (y2-y1)*y0;
Min1 = min(Min1,tmp1);
Max1 = max(Max1,tmp1);
Min2 = min(Min2,tmp2);
Max2 = max(Max2,tmp2);
}
double a = y2-y1;
double b = x2-x1;
ans = min(ans,(Max1-Min1)*(Max2-Min2)/(a*a+b*b));
}
printf("Case #%d:\n%.0lf\n",iCase,ans);
}

return 0;
}