poj 2398 Toy Storage 計算幾何+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

這題在poj 2318的基礎上有些變化,首先是輸入的直線不是從左到右順序輸入的,要自己排序,其次,輸出要按出現次數排序,而不是按區間個數排(剛開始沒看清題,直接看的樣例,其實還想複雜了,但是實現了,wa了兩發,再看了下輸出要求,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)//叉積小於0說明點在直線左側
                    right=mid-1;
                else left=mid+1;
            }
            c[left]++;
        }//jug[i]代表區間內點有i個的 區間 的個數
        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;
}