poj 2318

题意：给出矩形的左上和右下的坐标，在矩形中有n个木棒，木棒之间不会相交，然后给出木棒上下端点的横坐标，接着有m个玩具，给出玩具的坐标。输出木棒围成的区域中有玩具的个数。

思路：由于数据范围大，所以对有序的木棒二分，叉积判断点在木棒的哪端。

#include<iostream>#include<cstdio>using namespace std;const int maxn = 5010;struct point {    double x , y;    };double Xmult(point a , point b , point c) {    return (b.x-a.x)*(c.y-a.y) - (b.y-a.y)*(c.x-a.x);}point P[maxn] , Pu[maxn] , Pd[maxn];int  sum[maxn];int n , m , k;void Bin(point X) {    int l , r , ans;    double tmp;    l = 0; r = n+1;    while(l <= r) {        int mid = (l+r)/2;        tmp = Xmult(X , Pu[mid] , Pd[mid]);        if(tmp < 0) {            ans = mid;            //printf("ans = %d\n",ans);            r = mid-1;            } else l = mid+1;    }    //printf("id = %d\n",ans);    sum[ans-1] ++;}int main() {    double x1 , y1 , x2 , y2;    int i , j;    while(~scanf("%d",&n)) {        if(!n) break;        scanf("%d%lf%lf%lf%lf",&m,&x1,&y1,&x2,&y2);        for(i = 1 ; i <= n ; i ++) {            scanf("%lf%lf",&Pu[i].x,&Pd[i].x);            Pu[i].y = y1;            Pd[i].y = y2;        }        Pu[0].x = x1;Pu[0].y=y1;        Pd[0].x = x1;Pd[0].y=y2;        Pu[n+1].x = x2;Pu[n+1].y=y1;        Pd[n+1].x = x2;Pd[n+1].y=y2;        for(i = 0 ; i < m ; i ++) scanf("%lf%lf",&P[i].x,&P[i].y);        k  = 0;        memset(sum , 0 , sizeof(sum));        for(i = 0 ; i < m ; i ++) Bin(P[i]);        for(i = 0 ; i <= n ; i ++) printf("%d: %d\n",i,sum[i]);        printf("\n");    }    }