POJ 2318 TOYS （二分+折线拐向）

TOYS

Time Limit: 2000MS
Memory Limit: 65536KTotal Submissions: 15181
Accepted: 7324

Description

Calculate the number of toys that land in each bin of a partitioned toy box. 
Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them. They gave John a rectangular box to put his toys in, but John is rebellious and obeys his parents by simply throwing his toys into the box. All the toys get mixed up, and it is impossible for John to find his favorite toys. 

John's parents came up with the following idea. They put cardboard partitions into the box. Even if John keeps throwing his toys into the box, at least toys that get thrown into different bins stay separated. The following diagram shows a top view of an example toy box. 

For this problem, you are asked to determine how many toys fall into each partition as John throws them into the toy box.

Input

The input file contains one or more problems. The first line of a problem consists of six integers, n m x1 y1 x2 y2. The number of cardboard partitions is n (0 < n <= 5000) and the number of toys is m (0 < m <= 5000). The coordinates of the upper-left corner and the lower-right corner of the box are (x1,y1) and (x2,y2), respectively. The following n lines contain two integers per line, Ui Li, indicating that the ends of the i-th cardboard partition is at the coordinates (Ui,y1) and (Li,y2). You may assume that the cardboard partitions do not intersect each other and that they are specified in sorted order from left to right. The next m lines contain two integers per line, Xj Yj specifying where the j-th toy has landed in the box. The order of the toy locations is random. You may assume that no toy will land exactly on a cardboard partition or outside the boundary of the box. The input is terminated by a line consisting of a single 0.

Output

The output for each problem will be one line for each separate bin in the toy box. For each bin, print its bin number, followed by a colon and one space, followed by the number of toys thrown into that bin. Bins are numbered from 0 (the leftmost bin) to n (the rightmost bin). Separate the output of different problems by a single blank line.

Sample Input

5 6 0 10 60 03 14 36 810 1015 301 52 12 85 540 107 94 10 0 10 100 020 2040 4060 6080 80 5 1015 1025 1035 1045 1055 1065 1075 1085 1095 100

Sample Output

0: 21: 12: 13: 14: 05: 10: 21: 22: 23: 24: 2

Hint

As the example illustrates, toys that fall on the boundary of the box are "in" the box.

原题链接：POJ 2318 TOYS

题目大意：

 给你两个数n,m，n表示有多少条线，m表示有多少个玩具，然后给你两个坐标，为长方形的左上角和右下角。然后n行，每行两个数，表示线段的两头的x坐标，接着m行数，表示玩具的坐标。问，在线段隔开的每个区域里有多少个玩具。

主要用到的知识点就是线段的拐向和二分。

折线段的拐向判断：

折线段的拐向判断方法可以直接由矢量叉积的性质推出。对于有公共端点的线段p0p1和p1p2，通过计算(p2 - p0) × (p1 - p0)的符号便可以确定折线段的拐向：

若(p2 - p0) × (p1 - p0) > 0,则p0p1在p1点拐向右侧后得到p1p2。

若(p2 - p0) × (p1 - p0) < 0,则p0p1在p1点拐向左侧后得到p1p2。

若(p2 - p0) × (p1 - p0) = 0,则p0、p1、p2三点共线。

以下为AC代码

#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>#include <queue>#include <map>#include <vector>#include <set>#include <string>#include <math.h>using namespace std;struct Point{    int x,y;    Point(){}    Point(int _x,int _y)    {        x = _x;y = _y;    }    Point operator -(const Point &b)const    {        return Point(x - b.x,y - b.y);    }    int operator *(const Point &b)const    {        return x*b.x + y*b.y;    }    int operator ^(const Point &b)const    {        return x*b.y - y*b.x;    }};struct Line{    Point s,e;    Line(){}    Line(Point _s,Point _e)        {        s = _s;e = _e;    }};int xmult(Point p0,Point p1,Point p2) //计算p0p1 X p0p2{    return (p1-p0)^(p2-p0);}const int MAXN = 5050;Line line[MAXN];int ans[MAXN];int main(){    int n,m,x1,y1,x2,y2;    bool first = true;    while(scanf("%d",&n) == 1 && n)    {        if(first)first = false;        else printf("\n");        scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);        int Ui,Li;        for(int i = 0;i < n;i++)        {            scanf("%d%d",&Ui,&Li);            line[i] = Line(Point(Ui,y1),Point(Li,y2));        }        line[n] = Line(Point(x2,y1),Point(x2,y2));        int x,y;        Point p;        memset(ans,0,sizeof(ans));        while( m-- )        {            scanf("%d%d",&x,&y);            p = Point(x,y);            int l = 0,r = n;            int tmp;            while( l <= r)            {                int mid = (l + r)/2;                if(xmult(p,line[mid].s,line[mid].e) < 0)                {                    tmp = mid;                    r = mid - 1;                }                else l = mid + 1;            }            ans[tmp]++;        }        for(int i = 0; i <= n;i++)            printf("%d: %d\n",i,ans[i]);    }    return 0;}