poj2318

题意：一个矩形箱子，左上角与右下角的坐标给出，里面有n块板把箱子里的空间分隔成许多个分区，给出这些板在上边的x坐标、下边的x坐标，以及一堆玩具的坐标，求这些分区里的玩具数目。

分析：记玩具在点p0，某块板的上边点是p1，下边点是p2，p2p1(向量)×p2p0>0表示p0在p1p2的左面，<0表示在右面。接下来就是用二分法找出每个点所在的分区。

叉积+二分查找

#include<cstdio>
#define vector point
struct point
{
    int x,y;
    point(int xx,int yy)
    {
        x = xx;
        y = yy;
    }
    point operator - (const point& s)
    {
        return point(x - s.x, y - s.y);
    }
};
int cross_product(vector v1,vector v2)
{
    return v1.x * v2.y - v1.y * v2.x;
}
int x1,y1,x2,y2;
int n,m,board[5002][2],counter[5001];
int sort(int x,int y)
{
    int left = 0, right = n + 1,mid;
    while(right - left > 1)
    {
        mid = (left + right)/2;
        vector v1 = vector(point(board[mid][0],y1) - point(board[mid][1],y2)), v2 = vector(point(x,y) - point(board[mid][1],y2));
        /*printf("up = (%d,%d),down = (%d,%d), this.x = %d, this.y = %d\n",board[mid][0],y1,board[mid][1],y2,x,y);
        printf(" l = %d, r = %d, m = %d, (%d,%d)X(%d,%d) = %d\n",left,right,mid,v1.x,v1.y,v2.x,v2.y,cross_product(v1,v2));*/
        if(cross_product(vector(point(board[mid][0],y1) - point(board[mid][1],y2)),vector(point(x,y) - point(board[mid][1],y2))) < 0)//在右邊
            left = mid;
        else
            right = mid;
    }
    //printf("left = %d\n",left);
    return left;
}
int main()
{
    bool flag = true;
    while(scanf("%d",&n) && n)
    {
        scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);
        for(int i = 1;i <= n;i++)
        {
            scanf("%d%d",&board[i][0],&board[i][1]);
            counter[i] = 0;
        }
        counter[0] = 0;
        board[0][0] = board[0][1] = x1;
        board[n+1][0] = board[n+1][1] = x2;
        int x,y;
        for(int i = 0;i < m;i++)
        {
            scanf("%d%d",&x,&y);
            counter[sort(x,y)]++;
        }
        if(!flag)
            printf("\n");
        else
            flag = false;
        for(int i = 0;i <= n;i++)
            printf("%d: %d\n",i,counter[i]);
    }
    return 0;
}