poj 2318 TOYS (叉积)

http://poj.org/problem?id=2318

TOYS

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 13061		Accepted: 6298

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.

给定一个如上的长方形箱子，中间有n条线段，将其分为n+1个区域，给定m个玩具的坐标，统计每个区域中的玩具个数

思路：二分

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cmath>

using namespace std;

#define N 5110
#define INF 0x3f3f3f3f
#define MOD
#define met(a, b) memset(a, b, sizeof(a))

typedef long long LL;

using namespace std;

struct point
{
    int x, y;
};

struct line
{
    point down, up;
};

int ans[N];
line p[N];

int xmult (point p1, point p2, point p0)
{
    return (p1.x-p0.x) * (p2.y-p0.y) - (p2.x-p0.x) * (p1.y-p0.y);
}

void Search (point Q, int n)
{
    int l = 1, r = n;

    while (l < r)
    {
        int mid = (l + r) >> 1;

        if (xmult (Q, p[mid].up, p[mid].down) > 0) l = mid+1;///点在右端
        else r = mid;///点在左端
    }

    if (xmult (p[l].up, p[l].down, Q) < 0) ans[l]++;
    else ans[l+1]++;
}

int main ()
{
    int n, m, x1, y1, x2, y2;

    while (scanf ("%d", &n), n)
    {
        point Q;
        met (p, 0);
        met (ans, 0);

        scanf ("%d %d %d %d %d", &m, &x1, &y1, &x2, &y2);

        for (int i=1; i<=n; i++)
        {
            int u, l;
            scanf ("%d %d", &u, &l);
            p[i].up = (point) {u, y1};
            p[i].down = (point) {l, y2};
        }

        for (int i=0; i<m; i++)
        {
            scanf ("%d %d", &Q.x, &Q.y);
            Search (Q, n);
        }

        int nCase = 0;
        for (int i=1; i<=n+1; i++)
            printf ("%d: %d\n", nCase++, ans[i]);
        puts ("");
    }
    return 0;
}