POJ 2398.Toy Storage

227


题目



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Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box to put his toys in. Unfortunately, Reza 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 Reza to find his favorite toys anymore.

Reza's parents came up with the following idea. They put cardboard partitions into the box. Even if Reza keeps throwing his toys into the box, at least toys that get thrown into different partitions stay separate. The box looks like this from the top:

We want for each positive integer t, such that there exists a partition with t toys, determine how many partitions have t, toys.


The input consists of a number of cases. The first line consists of six integers n, m, x1, y1, x2, y2. The number of cardboards to form the partitions is n (0 < n <= 1000) and the number of toys is given in m (0 < m <= 1000). 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 each consists of two integers Ui Li, indicating that the ends of the ith cardboard is at the coordinates (Ui, y1) and (Li, y2). You may assume that the cardboards do not intersect with each other. The next m lines each consists of two integers Xi Yi specifying where the ith toy has landed in the box. You may assume that no toy will land on a cardboard.

A line consisting of a single 0 terminates the input.


For each box, first provide a header stating "Box" on a line of its own. After that, there will be one line of output per count (t > 0) of toys in a partition. The value t will be followed by a colon and a space, followed the number of partitions containing t toys. Output will be sorted in ascending order of t for each box.


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


Box
2: 5
Box
1: 4
2: 1




题解



POJ 2318.TOYS 一样
区别如下:
  • 木板线段需要排序
  • 输出格式改变


代码


点击显/隐代码
#include <algorithm>
#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], ans[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);
}

bool compare(Vector a, Vector b) { return a.real() < b.real(); }

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);

        sort(P1 + 1, P1 + 1 + n, compare);
        sort(P2 + 1, P2 + 1 + n, compare);
        // 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))]++;
        }

        memset(ans, 0, sizeof(ans));
        for (int i = 0; i <= n; ++i)
            ans[cnt[i]]++;

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