poj 2398 Toy Storage 计算几何+二分

Toy Storage

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5296		Accepted: 3136

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 0
20 20
80 80
60 60
40 40
5 10
15 10
95 10
25 10
65 10
75 10
35 10
45 10
55 10
85 10
5 6 0 10 60 0
4 3
15 30
3 1
6 8
10 10
2 1
2 8
1 5
5 5
40 10
7 9
0

Sample Output

Box
2: 5
Box
1: 4
2: 1

这题在poj 2318的基础上有些变化,首先是输入的直线不是从左到右顺序输入的,要自己排序,其次,输出要按出现次数排序,而不是按区间个数排(刚开始没看清题,直接看的样例,其实还想复杂了,但是实现了,wa了两发,再看了下输出要求,1A)

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
struct Point
{
    int x,y;
    Point (int a=0,int b=0):x(a),y(b) {}
};

inline int  multiply(Point sp,Point  ep,Point op)
{
    return((sp.x-op.x)*(ep.y-op.y)-(ep.x-op.x)*(sp.y-op.y));
}

bool cmp(const Point a,const Point b)
{
    return a.x<b.x;
}

Point a[1001],b[1001];
int c[1001],jug[1001];
int main()
{
    int N,M,x1,x2,y1,y2,n,m,mid,i;
    while(~scanf("%d",&N))
    {
        if(N<=0) break;
        for(i=0; i<1001; i++)
        {
            c[i]=jug[i]=0;
        }

        scanf("%d %d %d %d %d",&M,&x1,&y1,&x2,&y2);
        for(i=0; i<N; i++)
        {
            scanf("%d %d",&b[i].x,&b[i].y);
            //(b[i].x,y1) (b[i].y,y2);
        }
        sort(b,b+N,cmp);

        for(i=0; i<M; i++)//toys
            scanf("%d %d",&a[i].x,&a[i].y);
        for(i=0; i<M; i++)
        {
            int left=0,right=N-1;
            while(left<=right)
            {
                mid=(left+right)/2;
                if(multiply(Point(b[mid].x,y1),Point(b[mid].y,y2),a[i])<0)//叉积小于0说明点在直线左侧
                    right=mid-1;
                else left=mid+1;
            }
            c[left]++;
        }//jug[i]代表区间内点有i个的 区间 的个数
        for(i=0; i<=N; i++)
            jug[c[i]]++;
        printf("Box\n");
        for(i=1; i<=1000; i++)
            if(jug[i]!=0)
                printf("%d: %d\n",i,jug[i]);
    }
    return 0;
}