[kuangbin] 专题13 基础计算几何 题解 + 总结

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[kuangbin] 专题7 线段树 题解 + 总结：https://www.cnblogs.com/RioTian/p/13413897.html

kuangbin专题十二 基础DP1 题解+总结：https://www.cnblogs.com/RioTian/p/13110438.html

kuangbin专题六 最小生成树 题解+总结：https://www.cnblogs.com/RioTian/p/13380764.html

[kuangbin]专题九 连通图 题解+总结 ： https://www.cnblogs.com/RioTian/p/13395039.html

计算几何分类blog：https://www.cnblogs.com/RioTian/category/1852545.html

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总结

1、TOYS POJ - 2318

题目链接： Click Here

题目大意： 有一个方盒子 有N个板隔开 分成N+1个区域
又给了M个玩具的坐标 问你每个区域内（不能恰好在区域内）的玩具有几个（忽略玩具体积）

解题思路： 每相邻的两个板看成两个向量 分别求其与其中一点和玩具坐标的叉积 如果两叉积的乘积<0 就说明这个玩具坐标点在一个板的右边 一个板的左边

因为数据量比较大 可以把板和方盒的两边记录下来 然后二分 道理是一样的

AC代码：

Code

#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#define ms(a, b) memset(a, b, sizeof a)
using namespace std;
typedef long long ll;
const int N = 5000 + 10;
int num[N], x[N], y[N];
struct node {
    double x, y;
} pu[N], pl[N], q;
int n, m, f;
int mul(node p1, node p2, node p3) {
    return (p2.x - p1.x) * (p3.y - p1.y) - (p3.x - p1.x) * (p2.y - p1.y);
}
void Bisearch(node x) {
    int l = 0, r = n + 1, ans;
    while (l <= r) {
        int mid = (l + r) >> 1;
        if (mul(x, pu[mid], pl[mid]) < 0)
            ans = mid, r = mid - 1;
        else
            l = mid + 1;
    }
    num[ans - 1]++;
}
int main() {
    // freopen("in.txt", "r", stdin);
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int x1, x2, y1, y2;
    while (cin >> n) {
        if (n == 0) break;
        ms(num, 0);
        cin >> m >> x1 >> y1 >> x2 >> y2;
        // scanf("%d %d %d %d %d", &m, &x1, &y1, &x2, &y2);
        pu[0].x = x1, pu[0].y = y1;
        pl[0].x = x1, pl[0].y = y2;
        for (int i = 1; i <= n; ++i) {
            // scanf("%d %d", &pu[i].x, &pl[i].x);
            cin >> pu[i].x >> pl[i].x;
            pu[i].y = y1, pl[i].y = y2;
        }
        pu[n + 1].x = x2, pu[n + 1].y = y1;
        pl[n + 1].x = x2, pl[n + 1].y = y2;
        for (int i = 0; i < m; i++) {
            // scanf("%d %d", &q.x, &q.y);
            cin >> q.x >> q.y;
            Bisearch(q);
        }
        if (f == 1) printf("
");
        f = 1;
        for (int i = 0; i <= n; i++) printf("%d: %d
", i, num[i]);
    }
}

另一种写法

Code

#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
using namespace std;
typedef long long ll;
typedef double db;
const int maxn = 5e3 + 5;
const db eps = 1e-10;
int dcmp(db x) {
    if(fabs(x) < eps) {
        return 0;
    }
    return x > 0? 1: -1;
}
struct Point {
    double x, y;
    Point(double xx = 0, double yy = 0) : x(xx), y(yy) {}
    void input() {
        scanf("%lf%lf", &x, &y);
    }
    Point operator-(const Point a) {
        return Point(x - a.x, y - a.y);
    }
    db cross(const Point a) {
        return x * a.y - y * a.x;
    }
};
typedef Point Vector;
struct Line {
    Point s, e;
    Line() {}
    Line(Point s, Point e) : s(s), e(e) {}
    int toLeftTest(Point p) {
        if((e - s).cross(p - s) > 0) return 1;
        else if((e - s).cross(p - s) < 0) return -1;
        return 0;
    }
};
int n, m;
Line line[maxn];
Point point[maxn];
int ans[maxn];
void solve() {
    for(int i = 1; i <= m; ++i) {
        int l = 0, r = n + 1;
        int mid = (l + r) >> 1;
        while(l + 1 < r) {
            if(line[mid].toLeftTest(point[i]) > 0) {
                r = mid;
            } else {
                l = mid;
            }
            mid = (l + r) >> 1;
        }
        ans[l]++;
    }
    for(int i = 0; i <= n; ++i) {
        printf("%d: %d
", i, ans[i]);
    }
    printf("
");
}
int main() {
    while(scanf("%d", &n) != EOF && n) {
        memset(ans, 0, sizeof(ans));
        scanf("%d", &m);
        db xl, yl, xr, yr;
        scanf("%lf%lf%lf%lf", &xl, &yl, &xr, &yr);
        line[0].s = Point(xl, yr);
        line[0].e = Point(xl, yl);
        line[n + 1].s = Point(xr, yr);
        line[n + 1].e = Point(xr, yl);
        for(int i = 1; i <= n; ++i) {
            db u, l;
            scanf("%lf%lf", &u, &l);
            line[i].s = Point(l, yr);
            line[i].e = Point(u, yl);
        }
        for(int i = 1; i <= m; ++i) {
            scanf("%lf%lf", &point[i].x, &point[i].y);
        }
        solve();
    }
    return 0;
}

2、Toy Storage POJ - 2398

题目链接：Click Here

与 POJ 2318 类似，不同的是这次给定的线段是乱序的，因此先要排序。输出也不同，改一下即可。

Code

// Author : RioTian
// Time : 20/10/21
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
using namespace std;
typedef long long ll;
typedef double db;
const int maxn = 5e3 + 5;
const db eps = 1e-10;
int dcmp(db x) {
    if (fabs(x) < eps) {
        return 0;
    }
    return x > 0 ? 1 : -1;
}
struct Point {
    double x, y;
    Point(double xx = 0, double yy = 0) : x(xx), y(yy) {}
    void input() { scanf("%lf%lf", &x, &y); }
    bool operator<(const Point &a) const {
        return (!dcmp(x - a.x)) ? dcmp(y - a.y) < 0 : x < a.x;
    }
    Point operator-(const Point a) { return Point(x - a.x, y - a.y); }
    db cross(const Point a) { return x * a.y - y * a.x; }
};
typedef Point Vector;
struct Line {
    Point s, e;
    Line() {}
    Line(Point s, Point e) : s(s), e(e) {}
    bool operator<(const Line &a) const { return s < a.s; }
    int toLeftTest(Point p) {
        if ((e - s).cross(p - s) > 0)
            return 1;
        else if ((e - s).cross(p - s) < 0)
            return -1;
        return 0;
    }
};
int cmp(Line a, Line b) { return a.s < b.s; }
int n, m;
Line line[maxn];
Point point[maxn];
int ans[maxn];
int cnt[maxn];
void solve() {
    for (int i = 1; i <= m; ++i) {
        int l = 0, r = n + 1;
        int mid = (l + r) >> 1;
        while (l + 1 < r) {
            if (line[mid].toLeftTest(point[i]) > 0) {
                r = mid;
            } else {
                l = mid;
            }
            mid = (l + r) >> 1;
        }
        ans[l]++;
    }
    int max_size = 0;
    for (int i = 0; i <= n; ++i) {
        if (ans[i]) {
            cnt[ans[i]]++;
            max_size = max(max_size, ans[i]);
        }
    }
    printf("Box
");
    for (int i = 1; i <= max_size; ++i) {
        if (cnt[i]) {
            printf("%d: %d
", i, cnt[i]);
        }
    }
}
int main() {
    while (scanf("%d", &n) != EOF && n) {
        memset(ans, 0, sizeof(ans));
        memset(cnt, 0, sizeof(cnt));
        scanf("%d", &m);
        db xl, yl, xr, yr;
        scanf("%lf%lf%lf%lf", &xl, &yl, &xr, &yr);
        line[0].s = Point(xl, yr);
        line[0].e = Point(xl, yl);
        line[n + 1].s = Point(xr, yr);
        line[n + 1].e = Point(xr, yl);
        for (int i = 1; i <= n; ++i) {
            db u, l;
            scanf("%lf%lf", &u, &l);
            line[i].s = Point(l, yr);
            line[i].e = Point(u, yl);
        }
        sort(line + 1, line + 1 + n);  // 要对线段排序
        for (int i = 1; i <= m; ++i) {
            scanf("%lf%lf", &point[i].x, &point[i].y);
        }
        solve();
    }
    return 0;
}