[代码]ACM-ICPC 2012 Regionals Asia - Jinhua D Crazy Tank / HDU 4445

Abstract

ACM-ICPC 2012 Regionals Asia - Jinhua D Crazy Tank / HDU 4445

扫描线

Source

http://acm.hdu.edu.cn/showproblem.php?pid=4445

Solution

很多人是水过的，我贴个正经的扫描线的吧。

Code

#include <cassert>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <iostream>

#include <cmath>

using namespace std;



const double PI = acos(-1.0);

const double EPS = 1e-8;



inline int sgn(double x) {

    if (fabs(x)<EPS) return 0;

    return x>0?1:-1;

}



int N, M;

double l1, r1, l2, r2;

double v[222];

double g=9.8, h;



double calc(double v, double a) {

    double vx = v*cos(a);

    double vy = v*sin(a);

    double delta = vy*vy+2*g*h;

    delta = max(0.0, delta);

    double t = (vy+sqrt(delta))/g;

    return vx*t;

}



double tri(double l, double r, double v, double &d) {

    int cnt = 100;

    while (cnt--) {

        double m1 = (l+r)/2;

        double m2 = (m1+r)/2;

        if (calc(v, m1)>calc(v, m2)) r = m2;

        else l = m1;

    }

    d = calc(v, l);

    return l;

}



double bin(double l, double r, double d, double v, bool up) {

    int cnt = 100;

    while (cnt--) {

        double m = (l+r)/2;

        if ((calc(v, m)<d)^up) l = m;

        else r = m;

    }

    return l;

}



struct se {

    double a;

    bool in;

    int cnt;

    se() {}

    se(double x, bool y, int z): a(x), in(y), cnt(z) {}

    bool operator<(const se &rhs) const {

        if (sgn(a-rhs.a)==0) {

            return in<rhs.in;

        }

        return a<rhs.a;

    }

}e[9999];



void ae(double a, bool in, int cnt) {

    e[M++] = se(a, in, cnt);

}



void solve(double l, double r, double d, double v, double t, int cnt) {

    if (sgn(d-l)<0) return;

    if (sgn(d-r)<0) {

        double hmr = bin(-PI/2, t, l, v, 0);

        double mdk = bin(t, PI/2, l, v, 1);

        if (sgn(hmr-mdk)>0) while (1) puts("");

        ae(hmr, 0, cnt); ae(mdk, 1, -cnt);

        return;

    }

    double s1 = bin(-PI/2, t, l, v, 0);

    double t1 = bin(-PI/2, t, r, v, 0);

    double s2 = bin(t, PI/2, r, v, 1);

    double t2 = bin(t, PI/2, l, v, 1);

    if (sgn(s1-t1)>0) while (1) puts("");

    if (sgn(s2-t2)>0) while (1) puts("");

    if (sgn(t1-s2)>=0) {

        ae(s1, 0, cnt); ae(t2, 1, -cnt);

    }

    else {

            ae(s1, 0, cnt); ae(t1, 1, -cnt);

            ae(s2, 0, cnt); ae(t2, 1, -cnt);

    }

}



char s[111];



int main() {

    int i, j, k;

    while (cin>>N>>h>>l1>>r1>>l2>>r2, N) {

        M = 0;

        for (i = 0; i < N; ++i) {

            cin>>v[i];

            double d, t;

            t = tri(-PI/2, PI/2, v[i], d);

            solve(l1, r1, d, v[i], t, 1);

            solve(l2, r2, d, v[i], t, -998);

        }

        sort(e, e+M);

        int ans = 0;

        int cnt = 0;

        for (i = 0; i < M; ++i) {

            cnt += e[i].cnt;

            if (cnt>N) while (1) puts("");

            ans = max(ans, cnt);

        }

        printf("%d\n", ans);

    }

    return 0;

}