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Problem J: Saint John Festival

\[ Time Limit: 1 s \quad Memory Limit: 256 MiB \]

题意

给出\(n\)个大点，和\(m\)个小点，然后问有多少个小点可以在任意一个\(3\)个大点组成的三角形内。

思路

很明显只要对大点求凸包，然后判断有多少个在凸包里的小点就可以了，但是判断点在凸包内如果用\(O(N)\)的方法会\(TLE\)，需要进行二分。
我求出的是逆时针的凸包，然后定下一个端点\(p[1]\)，寻找另外两个端点\(p[id]\)和\(p[id+1]\)，根据查询的点在\(p[1]-p[id]\)这条直线右侧或者在\(p[1]-p[id+1]\)这个点左侧来二分范围，如果在\(p[1]-p[id]\)和\(p[1]-p[id+1]\)之间，那么在判断是否在\(p[id]-p[id+1]\)左侧来判断是否在凸包内。

/***************************************************************
    > File Name    : J.cpp
    > Author       : Jiaaaaaaaqi
    > Created Time : 2019年05月06日 星期一 18时14分21秒
 ***************************************************************/

#include <map>
#include <set>
#include <list>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <cfloat>
#include <string>
#include <vector>
#include <cstdio>
#include <bitset>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define  lowbit(x)  x & (-x)
#define  mes(a, b)  memset(a, b, sizeof a)
#define  fi         first
#define  se         second
#define  pii        pair<int, int>
#define  INOPEN     freopen("in.txt", "r", stdin)
#define  OUTOPEN    freopen("out.txt", "w", stdout)

typedef unsigned long long int ull;
typedef long long int ll;
const int    maxn = 5e4 + 10;
const int    maxm = 1e5 + 10;
const ll     mod  = 1e9 + 7;
const ll     INF  = 1e18 + 100;
const int    inf  = 0x3f3f3f3f;
const double pi   = acos(-1.0);
const double eps  = 1e-8;
using namespace std;

int n, m;
int cas, tol, T;
int sgn(double x) {
    if(fabs(x) <= eps)  return 0;
    else    return x>0 ? 1 : -1;
}
struct Point {
    double x, y;
    Point() {}
    inline Point(double _x, double _y) {
        x = _x, y = _y;
    }
    inline Point operator - (Point a) const {
        return Point(x-a.x, y-a.y);
    }
    inline double operator ^ (Point a) const {
        return x*a.y - y*a.x;
    }
    inline double distance(Point p) const {
        return hypot(x-p.x, y-p.y);
    }
    inline bool operator < (Point a) const {
        return sgn(y-a.y)==0 ? sgn(x-a.x)<0 : y<a.y;
    }
    inline bool operator == (Point a) const {
        return sgn(x-a.x)==0 && sgn(y-a.y)==0;
    }
    inline double operator * (Point a) const {
        return x*a.x + y*a.y;
    }
};
struct Line {
    Point s, e;
    Line() {}
    Line(Point _s, Point _e) {
        s = _s, e = _e;
    }
    inline bool pointseg(Point p) {
        return sgn((p-s)^(e-s)) == 0 && sgn((p-s)*(p-e)) <=0;
    }
};
struct Polygon {
    int n;
    Point p[maxn];
    Line l[maxn];
    inline void add(Point q) {
        p[++n] = q;
    }
    struct cmp {
        Point p;
        cmp(Point _p) {
            p = _p;
        }
        bool operator() (Point _a, Point _b) const {
            Point a = _a, b = _b;
            int d = sgn((a-p)^(b-p));
            if(d == 0) {
                return sgn(a.distance(p) - b.distance(p)) < 0;
            } else {
                return d>0;
            }
        }
    };
    void norm() {
        int id = 1;
        for(int i=2; i<=n; ++i) {
            if(p[i] < p[id])
                id = i;
        }
        swap(p[id], p[1]);
        sort(p+1, p+1+n, cmp(p[1]));
    }
    void Graham(Polygon &convex) {
        norm();
        mes(convex.p, 0);
        int &top = convex.n = 0;
        if(n == 1) {
            convex.p[++top] = p[1];
        } else if(n == 2) {
            convex.p[++top] = p[1];
            convex.p[++top] = p[2];
            if(convex.p[1] == convex.p[2])  top--;
        } else {
            convex.p[++top] = p[1];
            convex.p[++top] = p[2];
            for(int i=3; i<=n; ++i) {
                while(top>1 && sgn((convex.p[top]-convex.p[top-1])^
                    (p[i]-convex.p[top-1])) <= 0)
                        top--;
                convex.p[++top] = p[i];
            }
            if(top == 2 && convex.p[1] == convex.p[2])
                top--;
        }
    }
    void getline() {
        for(int i=1; i<=n; ++i) {
            l[i] = Line(p[i], p[i%n+1]);
        }
    }
    int inconvex(Point s) {
        /*
        点和凸包的关系
        2   边上
        1   内部
        0   外部
        */
        Point p1 = p[1];
        Line l1 = Line(p[1], p[2]);
        Line l2 = Line(p[1], p[n]);
        if(l1.pointseg(s) || l2.pointseg(s))
            return 2;
        int l = 2, r = n-1;
        while(l<=r) {
            int mid = l+r>>1;
            int t1 = sgn((s-p1)^(p[mid]-p1));
            int t2 = sgn((s-p1)^(p[mid+1]-p1));
            if(t1 <= 0 && t2 >= 0) {
                int t3 = sgn((s-p[mid]) ^ (p[mid+1]-p[mid]));
                if(t3 < 0)  return 1;
                else if(t3 == 0)    return 2;
                return 0;
            }
            if(t1 > 0)  r = mid-1;
            else    l = mid+1;
        }
        return 0;
    }
} large, small, con;
inline int read() {
    int x = 0, f = 1;
    char s = getchar();
    while (s < '0' || s > '9') {
        if (s == '-')f = -1;
        s = getchar();
    }
    while (s >= '0' && s <= '9') {
        x = x * 10 + s - '0';
        s = getchar();
    }
    return x * f;
}
int main() {
    n = read();
    large.n = small.n = con.n = 0;
    int x, y;
    for(int i=1; i<=n; ++i) {
        x = read(), y =read();
        large.add(Point(1.0*x, 1.0*y));
    }
    m = read();
    for(int i=1; i<=m; ++i) {
        x = read(), y =read();
        small.add(Point(1.0*x, 1.0*y));
    }
    large.norm();
    large.Graham(con);
    int ans = 0;
    for(int i=1; i<=m; ++i) {
        if(con.inconvex(small.p[i])) {
            ans++;
        }
    }
    printf("%d\n", ans);
    return 0;
}

Saint John Festival Gym - 101128J (凸包二分)

标签：add   float   mib   sign   aaaaaa   line   for   point   需要   

原文地址：https://www.cnblogs.com/Jiaaaaaaaqi/p/10821232.html