poj 2318 TOYS —–叉积+二分

转自Scar的个人网站   heroscar.com

TOYS

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7587		Accepted: 3598

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 0
3 1
4 3
6 8
10 10
15 30
1 5
2 1
2 8
5 5
40 10
7 9
4 10 0 10 100 0
20 20
40 40
60 60
80 80
 5 10
15 10
25 10
35 10
45 10
55 10
65 10
75 10
85 10
95 10
0

Sample Output

0: 2
1: 1
2: 1
3: 1
4: 0
5: 1

0: 2
1: 2
2: 2
3: 2
4: 2

题目大意： 如上图，在一个方形的盒子里，用n条线段划分成 n+1个区域。  再给出 m个点。统计 各个区域 有多少个点。点不会超出盒子的边界，也不会刚好在两个区域的分界线上。
解题思路：
 值得注意的是题目说明了  边界线的给出 是按 从左到右给出的，就可以联想到单调区间查找 可以用二分查找 提高效率。
但是 我一开始做的时候 是 从左到右 以此看点是否在 此线段的左边，是的就肯定在 这个线段左边的区域。
也过了300+ms。但是用二分 优化后到了100+ms。 贴出 二分的代码。

Problem: 2318		User: 201150080226
Memory: 280K		Time: 125MS
Language: C++		Result: Accepted

#include<stdio.h>
#include<string.h>
const int Max=5500;
double p[Max][2];
int cout[Max];
double Cross(double x1,double y1,double x2,double y2) { return x1*y2 - x2* y1; } //叉积，计算几何的基础
int main()
{
//  freopen("in.txt","r",stdin);
    double x1,y1,x2,y2,x3,y3;
    int i,j,n,m;
    while(scanf("%d",&n)!=EOF&&n)
    {
        scanf("%d%lf%lf%lf%lf",&m,&x1,&y1,&x2,&y2);
        memset(cout,0,sizeof(int)*(n+1));
        double rx1, ry1 = y1-y2, rx2, ry2;
        p[0][0]=p[0][1]=x1;
        p[n+1][0]=p[n+1][1]=x2;
        for(i=1;i<=n;i++)
            scanf("%lf%lf",&p[i][0],&p[i][1]);
        for(i=0;i<m;i++)
        {
            scanf("%lf%lf",&x3,&y3);
            int mid,max=n+1,min=0;
            while(min<max-1)       //二分优化
            {
                mid=(max+min)/2;
                rx1=p[mid][0]-p[mid][1];
                rx2=x3-p[mid][1];
                ry2=y3-y2;
                if(Cross(rx1,ry1,rx2,ry2) > 0)
                    max=mid;
                else min=mid;
            }
            cout[min]++;
        }
        for(i=0;i<=n;i++)
            printf("%d: %d\n",i,cout[i]);
        printf("\n");
    }
    return 0;
}