poj 1032 && zoj 1265 Area

#include<stdio.h>#include<math.h>#include<iostream>#include<algorithm>using namespace std;#define N 102struct 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;}