POJ 2398 Toy Storage(计算几何)

最新推荐文章于 2019-07-23 23:38:56 发布

lab104_yifan

最新推荐文章于 2019-07-23 23:38:56 发布

阅读量732

点赞数 1

CC 4.0 BY-SA版权

分类专栏：计算几何

本文链接：https://blog.youkuaiyun.com/accelerator_/article/details/44305097

计算几何专栏收录该内容

35 篇文章

订阅专栏

本文介绍了一种使用叉积判断结合二分搜索的方法来解决几何问题，通过排序输入的线段并计算每个点所在的格子数。代码实现了输入线段范围和点坐标后，输出每个数量级别的格子数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

和POJ2318一样的方法，都是利用叉积判断+二分，不过这题要先排序，还有输出的是，每个数量的格子数

代码：

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

const int N = 1005;

int n, m, x1, y1, x2, y2;

struct Point {
    int x, y;
    Point() {}
    Point(int x, int y) {
        this->x = x;
        this->y = y;
    }
};

typedef Point Vector;

Vector operator - (Vector A, Vector B) {
    return Vector(A.x - B.x, A.y - B.y);
}

struct Seg {
    Point a, b;
    Seg() {}
    Seg(Point a, Point b) {
        this->a = a;
        this->b = b;
    }
} seg[N];

int ans[N];

int Cross(Vector A, Vector B) {return A.x * B.y - A.y * B.x;}

void gao(Point p) {
    int l = 0, r = n;
    while (l < r) {
        int mid = (l + r) / 2;
        if (Cross(seg[mid].a - p, seg[mid].b - p) < 0) r = mid;
        else l = mid + 1;
    }
    ans[l]++;
}

bool cmp(Seg a, Seg b) {
    return a.a.x < b.a.x;
}

int out[N];

int main() {
    while (~scanf("%d", &n) && n) {
        memset(ans, 0, sizeof(ans));
        scanf("%d%d%d%d%d", &m, &x1, &y1, &x2, &y2);
        int x, y;
        for (int i = 0; i < n; i++) {
            scanf("%d%d", &x, &y);
            seg[i] = Seg(Point(x, y1), Point(y, y2));
        }
        sort(seg, seg + n, cmp);
        for (int i = 0; i < m; i++) {
            scanf("%d%d", &x, &y);
            gao(Point(x, y));
        }
        memset(out, 0, sizeof(out));
        for (int i = 0; i <= n; i++)
            if (ans[i]) out[ans[i]]++;
        printf("Box\n");
        for (int i = 1; i <= 1000; i++)
            if (out[i]) printf("%d: %d\n", i, out[i]);
    }
    return 0;
}