POJ 2318 2398 计算几何

最新推荐文章于 2020-08-21 16:36:49 发布

茗_ming

最新推荐文章于 2020-08-21 16:36:49 发布

阅读量466

点赞数

CC 4.0 BY-SA版权

分类专栏： ACM_计算几何

本文链接：https://blog.youkuaiyun.com/livenls/article/details/12260309

ACM_计算几何专栏收录该内容

1 篇文章

订阅专栏

本文介绍了一种使用交叉乘积判断点是否位于两个线段之间的方法，并通过二分查找优化求解过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

求叉积根据叉积的正负，判断点在哪两个线段之间，二分即可

#include <cstdio>
#include <cstring>

struct Point
{
    int x, y;
    Point(int a = 0, int b = 0)
    {
        x = a; y = b;
    }
};

Point tp[5555], bt[5555];
int cnt[5555];
int n;

int cross(Point x, Point y, Point z)
{
    return (x.x - y.x) * (z.y - y.y) - (x.y - y.y) * (z.x - y.x);
}

void solve(Point x)
{
    int l, r , mid;
    l = 0; r = n;
    while (l <= r)
    {
        mid = (l + r) >> 1;
        if (cross(x, tp[mid], bt[mid]) < 0) l = mid + 1;
        else r = mid - 1;
    }
//    printf("%d\n", l);
    cnt[l]++;
}



int main()
{
    int m;
    Point lt, rb;
    int t = 0;
    while (scanf("%d%d%d%d%d%d", &n, &m, <.x, <.y, &rb.x, &rb.y) == 6)
    {
        if (t == 1) printf("\n");
        int i, j, k;
        Point s;
        memset(cnt, 0, sizeof(cnt));
        for (i = 0 ; i < n; ++i)
        {
            scanf("%d%d", &tp[i].x, &bt[i].x);
            tp[i].y = lt.y;
            bt[i].y = rb.y;
        }
        tp[n].x = rb.x; tp[n].y = lt.y;
        bt[n].x = rb.x; bt[n].y = rb.y;
        for (i = 0; i < m; ++i)
        {
            scanf("%d%d", &s.x, &s.y);
            solve(s);
        }
        for (i = 0; i <= n; ++i)
            printf("%d: %d\n", i, cnt[i]);
        t = 1;
//        printf("\n");
    }
}

2398 增加排序和需要对计数计数

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

struct Point
{
    int x, y;
    Point(int a = 0, int b = 0)
    {
        x = a; y = b;
    }
};

Point tp[5555], bt[5555];
int cnt[5555], cnt1[5555];
int n;

int cross(Point x, Point y, Point z)
{
    return (x.x - y.x) * (z.y - y.y) - (x.y - y.y) * (z.x - y.x);
}

void solve(Point x)
{
    int l, r , mid;
    l = 0; r = n;
    while (l <= r)
    {
        mid = (l + r) >> 1;
        if (cross(x, tp[mid], bt[mid]) < 0) l = mid + 1;
        else r = mid - 1;
    }
//    printf("%d\n", l);
    cnt[l]++;
}

inline bool cmp(Point x, Point y)
{
    return x.x < y.x;
}



int main()
{
    int m;
    Point lt, rb;
    while (scanf("%d%d%d%d%d%d", &n, &m, <.x, <.y, &rb.x, &rb.y) == 6)
    {
        int i, j, k;
        Point s;
        memset(cnt, 0, sizeof(cnt));
        memset(cnt1, 0, sizeof(cnt1));
        for (i = 0 ; i < n; ++i)
        {
            scanf("%d%d", &tp[i].x, &bt[i].x);
            tp[i].y = lt.y;
            bt[i].y = rb.y;
        }
        tp[n].x = rb.x; tp[n].y = lt.y;
        bt[n].x = rb.x; bt[n].y = rb.y;
        sort(tp, tp + n + 1, cmp);
        sort(bt, bt + n + 1, cmp);
        for (i = 0; i < m; ++i)
        {
            scanf("%d%d", &s.x, &s.y);
            solve(s);
        }
        for (i = 0; i <= n; ++i)
            if (cnt[i])
                cnt1[cnt[i]]++;
        printf("Box\n");
        for (i = 0; i <= n; ++i)
            if (cnt1[i]) printf("%d: %d\n", i, cnt1[i]);
    }
}