poj2398 几何运算（叉乘---判断是否在凸四边形内的判断（二分、排序））

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <cstring>#include <string.h>        //poj2398#include <queue>            //几何运算（叉乘---判断是否在凸四边形内的判断（二分））#define N 1010 using namespace std;struct zuobiao{int x1, y1, x2, y2;}p[N];int b, c, a[N], d[N];  //数组d的下标表示的是有n（下标）个点的有多少部分bool cmp(zuobiao a1, zuobiao b1)     //排序，使其递增{return a1.x2<b1.x2;}int check(int t)       //叉乘的运算x1*y2-x2*y1(叉乘不满足交换律，所以可以判断其在左还是右）{int x1, y1;x1=b-p[t].x2;y1=c-p[t].y2;return p[t].x1*y1-p[t].y1*x1;}int main(){int n, m, t, x1, y1, x2, y2, k, g, h, T=0;while(scanf("%d", &n)&&n){scanf("%d%d%d%d%d", &m, &x1, &y1, &x2, &y2);p[0].x1=0;p[0].y1=y1;p[0].x2=x1;p[0].y2=y2;for(t=1; t<=n; ++t){scanf("%d%d", &p[t].x1, &p[t].x2);p[t].y1=y1;p[t].y2=y2;p[t].x1=p[t].x1-p[t].x2;p[t].y1=p[t].y1-p[t].y2;}sort(p+1, p+n+1, cmp); memset(a, 0, sizeof(a));memset(d, 0, sizeof(d));for(t=0; t<m; ++t){scanf("%d%d", &b, &c);if(b<x1||b>x2||c>y1||c<y2)continue;if(check(1)>0){a[0]++;continue;}if(check(n)<0){a[n]++;continue;}int l=1, r=n, mid;h=0;   //h用来储存上一次变换符号后叉乘的值 while(l<=r)   //二分{mid=(l+r)>>1;k=check(mid);g=check(mid-1);if(k>0){if(g<0){a[mid-1]++;break;}if(h<0){r=mid;h=k;continue;}r=mid-1;h=k;}else if(k<0){if(h>0){l=mid;h=k;continue;}l=mid+1;h=k;}else if(k==0||g==0)break;}}for(t=h=0; t<=n; ++t){if(!d[a[t]])h++;d[a[t]]++;}printf("Box\n");for(t=1, k=0; t<1001; ++t){if(k==h)break;if(d[t]){printf("%d: %d\n", t, d[t]);k++;}}}return 0;}