【POJ - 2318 TOYS】 点与线的关系

A - TOYS

 

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

Hint

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

题意：有一个盒子用n块板分成了n+1块，现在已知各板的两端顶点坐标，已知盒子顶点的坐标，给出m个点坐标判断这些点处于哪些格子中，输出每个格子中点的个数。

分析：计算几何基础——叉积。判断点在挡板的左边还是右边，找到挡板的过程可以用二分处理。输入一个坐标（x, y）时，判断向量（x - uu[mid], y - y1）与向量（ x - ll[mid], y - y2）叉积的正负。正数则点落在板右侧，更新左端点l的值，负数则点落在板左侧，更新右端点r的值。

代码如下：

#include <map>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long

using namespace std;
const int MX = 5e3 + 5;
const int mod = 1e9 + 7;
const int INF = 2e9 + 5;

int uu[MX], ll[MX], num[MX];

int pan(int x1, int y1, int x2, int y2){
    return x1 * y2 - x2 * y1;
}

int main(){
    int n, m, x1, x2, y1, y2;
    while(~scanf("%d", &n), n){
        memset(num, 0, sizeof(num));
        scanf("%d%d%d%d%d", &m, &x1, &y1, &x2, &y2);
        for(int i = 1; i <= n; i++){
            scanf("%d%d", &uu[i], &ll[i]);
        }
        for(int i = 1; i <= m; i++){
            int x, y;
            scanf("%d%d", &x, &y);
            int l = 1;
            int r = n+1;
            int mid = (l+r) / 2;
            while(l < r){
                if(pan(x - uu[mid], y - y1, x - ll[mid], y - y2) > 0)   l = mid + 1;
                else r = mid;
                mid = (l + r) / 2;
            }
            //cout << mid << " " << l << " " << r << endl;
            num[mid-1]++;
        }
        for(int i = 0; i <= n; i++){
            printf("%d: %d\n", i, num[i]);
        }
        printf("\n");
    }
    return 0;
}