TOYS（计算几何基础+点与直线的位置关系）

题目链接 ： http://poj.org/problem?id=2318

题面：

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

Hint

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

 

思路：对于每一条分界线，首先进行排序（根据他的上面的x排和下面的排都是可以的），然后标号从0到n-1。然后问题就转换为现在访问的点会在哪条分界线的左端，若落在第i条直线的左端，那么它也就落在区域i内，确定所属区域后直接return。在最后面对第n个区域进行处理前面没被return的点。对于本题的关键就是处理点与直线的位置关系，而这个用向量的叉积进行处理即可。

代码实现如下：

#include 
#include 
#include 
using namespace std;

const int maxn = 5e3 + 7;
int n, m, x1, x2, y1, y2;
int num[maxn];


struct node {
    int x, y;
    bool operator < (const node &b) const {
        return x < b.x;
    }
} P[maxn], L[maxn], nw, nxt;  //P储存所访问的点（其实也可以不用数组的），L储存直线（对于直线此处的x为上方的x，y为下方的y）；

double dot(node a, node b) {
    return (a.x * b.y - a.y * b.x);
}

bool check(node p, node pp) {
    nw.x = p.x - pp.y, nw.y = p.y - y2, nxt.x = pp.x - pp.y, nxt.y = y1 - y2;
    if( dot(nw, nxt) < 0) {  //为负则说明当前访问的点在该直线的左端；
        return true;
    } else
        return false;
}

void Throw(node p) {
    for(int i = 0; i < n; i++) {
        if(check(p, L[i])) {
            num[i]++;
            return;
        }
    }
    num[n]++;
    return;
}

int main() {
    //freopen("in.txt", "r", stdin);
    while(~scanf("%d", &n) && n) {
        scanf("%d%d%d%d%d", &m, &x1, &y1, &x2, &y2);
        memset(num, 0, sizeof(num));
        for(int i = 0; i < n; i++) {
            scanf("%d%d", &L[i].x, &L[i].y);
        }
        sort(L, L + n);
        for(int i = 0; i < m; i++) {
            scanf("%d%d", &P[i].x, &P[i].y);
            Throw(P[i]);
        }
        for(int i = 0; i <= n; i++) {
            printf("%d: %d\n", i, num[i]);
        }
        printf("\n");
    }
    return 0;
}