POJ 2318.TOYS

559


题目



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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.


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.


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.


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


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

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




题解



使用向量判断点在直线哪一侧

如 判断点$P_0$在线段$P_1P_2$的哪一侧:
$vec {P1P2} times vec {P1P0}$
若该值大于$0$,则在顺时针方向,小于$0$在逆时针方向,等于$0$在线段上

注意判断的时候应该使用二分判断

写的时候初始化写错变量了,调了好久

代码


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#include <complex>
#include <cstdio>
#include <cstring>
using namespace std;

#define Log(format, ...) printf(format, ##__VA_ARGS__)

/* 向量模板 */
typedef complex<double> Vector;

int Cross(Vector a, Vector b) {
    return a.real() * b.imag() - a.imag() * b.real();
}
int Dot(Vector a, Vector b) {
    return a.real() * b.real() + a.imag() * b.imag();
}

const int maxn = 5005;
Vector P1[maxn];
Vector P2[maxn];
int cnt[maxn];

bool Could(Vector p, Vector L) { return (Cross(L, p) >= 0); }

int Division(int l, int r, Vector p) {
    // printf("%d %d (%.f,%.f)\n", l, r, p.real(), p.imag());
    if (r - l == 1)
        return l;

    int mid = (l + r) >> 1;
    if (Could(p - P1[mid], P2[mid] - P1[mid]))
        return Division(l, mid, p);
    else
        return Division(mid, r, p);
}

int main() {
    int n, m, x1, y1, x2, y2;
    while (scanf("%d%d%d%d%d%d", &n, &m, &x1, &y1, &x2, &y2), n != 0) {
        P1[0] = Vector(x1, y2);
        P2[0] = Vector(x1, y1);
        for (int i = 1; i <= n; ++i) {
            int u, l;
            scanf("%d%d", &u, &l);
            P2[i] = Vector(u, y1);
            P1[i] = Vector(l, y2);
        }
        P1[n + 1] = Vector(x2, y2);
        P2[n + 1] = Vector(x2, y1);

        // for (int i = 0; i <= n + 1; ++i) {
        //     printf("%d : (%.f,%.f) -> (%.f,%.f)\n", i, P1[i].real(),
        //            P1[i].imag(), P2[i].real(), P2[i].imag());
        // }

        memset(cnt, 0, sizeof(int) * (n + 5));
        for (int i = 0; i < m; ++i) {
            int x, y;
            scanf("%d%d", &x, &y);
            cnt[Division(0, n + 1, Vector(x, y))]++;
        }

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