poj 1032 && zoj 1265 Area

#include<stdio.h>
#include<math.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 102
struct point
{
    double x;
    double y;
}p[N],s[N];
double det(double x1,double y1,double x2,double y2)
{
    return x1*y2-x2*y1;
}

double area(point a,point b,point c)
{
    return det(a.x-c.x,a.y-c.y,b.x-c.x,b.y-c.y);
}

bool cmp1(point a,point b)      //按照纵坐标从小到大排序
{
    if(a.y==b.y)return a.x<b.x;
    return a.y<b.y;
}

bool cmp2(point a,point b)       //按照横坐标从小到大排序
{
    if(a.x==b.x)return a.y<b.y;
    return a.x<b.x;
}
double dot(point a,point b,point c)
{
    return (a.x-c.x)*(b.x-c.x)+(a.y-c.y)*(b.y-c.y);
}
int main()
{
    int t,cn=0;
    double x,y;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        cn++;
        scanf("%d",&n);
        for(int i=0;i<n;i++)     //输入移动的单位 并化为点
        {
            scanf("%lf%lf",&x,&y);
            if(i==0)
            {
                p[i].x=x;
                p[i].y=y;
                s[i]=p[i];
            }
            else
            {
                p[i].x=p[i-1].x+x;
                p[i].y=p[i-1].y+y;
                s[i]=p[i];
            }
        }
        double ans=0;
        for(int i=1;i<n-1;i++)   //求面积
        {
            ans+=area(p[i],p[(i+1)%n],p[0])/2.0;
        }
        double xmin,xmax,ymin,ymax;

        sort(s,s+n,cmp1);
        ymin=s[0].y;
        ymax=s[n-1].y;

        sort(s,s+n,cmp2);
        xmin=s[0].x;
        xmax=s[n-1].x;

        int ans1=n;
        for(double i=xmin;i<=xmax;i++)
        {
            for(double j=ymin;j<=ymax;j++)
            {
                p[n].x=i;
                p[n].y=j;
                for(int k=0;k<n;k++)
                {
                    if(area(p[(k+1)%n],p[n],p[k])==0&&dot(p[(k+1)%n],p[k],p[n])<0)
                    {
                        ans1++;
                        break;
                    }
                }
            }
        }
        if(cn==1)
        {
            printf("Scenario #%d:\n",cn);
            printf("%d %d %.1lf\n",(int)ans-ans1/2+1,ans1,ans);
        }
        else
        { 
            printf("Scenario #%d:\n",cn);
            printf("%d %d %.1lf\n",(int)ans-ans1/2+1,ans1,ans);
        }
        printf("\n");
    }
    return 0;
}