POJ 2398 Toy Storage(二分加折线拐向）

Toy Storage

Time Limit: 1000MS
Memory Limit: 65536KTotal Submissions: 5679
Accepted: 3383

Description

Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box to put his toys in. Unfortunately, Reza 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 Reza to find his favorite toys anymore. 
Reza's parents came up with the following idea. They put cardboard partitions into the box. Even if Reza keeps throwing his toys into the box, at least toys that get thrown into different partitions stay separate. The box looks like this from the top: 


We want for each positive integer t, such that there exists a partition with t toys, determine how many partitions have t, toys.

Input

The input consists of a number of cases. The first line consists of six integers n, m, x1, y1, x2, y2. The number of cardboards to form the partitions is n (0 < n <= 1000) and the number of toys is given in m (0 < m <= 1000). 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 each consists of two integers Ui Li, indicating that the ends of the ith cardboard is at the coordinates (Ui, y1) and (Li, y2). You may assume that the cardboards do not intersect with each other. The next m lines each consists of two integers Xi Yi specifying where the ith toy has landed in the box. You may assume that no toy will land on a cardboard. 

A line consisting of a single 0 terminates the input.

Output

For each box, first provide a header stating "Box" on a line of its own. After that, there will be one line of output per count (t > 0) of toys in a partition. The value t will be followed by a colon and a space, followed the number of partitions containing t toys. Output will be sorted in ascending order of t for each box.

Sample Input

4 10 0 10 100 020 2080 8060 6040 405 1015 1095 1025 1065 1075 1035 1045 1055 1085 105 6 0 10 60 04 315 303 16 810 102 12 81 55 540 107 90

Sample Output

Box2: 5Box1: 42: 1

Source

Tehran 2003 Preliminary

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，区间有相同玩具数的区间有多少个，玩具有多少个。

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。

这道题目和POJ 2318是一样的题目。POJ 2318 博客

需要注意的是，线段并不是按顺序输入的，所以输完线段要排序一下。

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;int num[1005];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;    }    bool friend operator < (Line a,Line b){    return a.e.x<b.e.x;    }};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;        memset(num,0,sizeof(num));        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));        sort(line,line+n);        int x,y;        Point p;        memset(ans,0,sizeof(ans));        while( m-- )        {            scanf("%d%d",&x,&y);            //if(x<=x1||x>=x2||y<=y2||y>=y1)continue;            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]++;//cout<<"tmp="<<tmp<<endl;        }        for(int i = 0; i <= n;i++){        if(ans[i]!=0)num[ans[i]]++;        }        printf("Box\n");        for(int i=0;i<=n;i++){        if(num[i]!=0)printf("%d: %d\n",i,num[i]);        }                    //printf("%d: %d\n",i,ans[i]);    }    return 0;}

原题连接：POJ 2398