HDU 4445 Crazy Tank

　　先反求可行的和不可行的出射角度区间，再扫描线求最大覆盖区间数。

// 2013-04-23 21:46:47  Accepted    4445    15MS    256K    2249 B  G++ Aros#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;const int MAXN = 200+5;const double g = 9.8, eps = 1e-8;int N, r[MAXN<<2], q[MAXN<<2];double H, L1, R1, L2, R2, V[MAXN];double x1, x2, x3, x4, x[MAXN<<2], y[MAXN<<2];bool calc(double V, double s, double &x, double &y){    if ((g*s*s/(V*V)-H)/sqrt(H*H+s*s)+eps < -1 || (g*s*s/(V*V)-H)/sqrt(H*H+s*s) > 1+eps)        return 0;    x = atan(s/H)-acos((g*s*s/(V*V)-H)/sqrt(H*H+s*s));    y = acos((g*s*s/(V*V)-H)/sqrt(H*H+s*s))+atan(s/H);    return 1;}bool cmpx(const int &a, const int &b){    if (abs(x[a]-x[b]) > eps)        return x[a]+eps < x[b];    else        return (a&1) < (b&1);}bool cmpy(const int &a, const int &b){    if (abs(y[a]-y[b]) > eps)        return y[a]+eps < y[b];    else        return (a&1) < (b&1);}int main(){    while (scanf("%d", &N) && N)    {        scanf("%lf%lf%lf%lf%lf", &H, &L1, &R1, &L2, &R2);        int c = 0, e = 0;        for (int i = 1; i <= N; i++)        {            scanf("%lf", &V[i]);            if (calc(V[i], L2, x1, x4))            {                if (calc(V[i], R2, x2, x3))                {                    r[c] = c, x[c++] = x1;                    r[c] = c, x[c++] = x2;                    r[c] = c, x[c++] = x3;                    r[c] = c, x[c++] = x4;                }                else                {                    r[c] = c, x[c++] = x1;                    r[c] = c, x[c++] = x4;                }            }            if (calc(V[i], L1, x1, x4))            {                if (calc(V[i], R1, x2, x3))                {                    q[e] = e, y[e++] = x1;                    q[e] = e, y[e++] = x2;                    q[e] = e, y[e++] = x3;                    q[e] = e, y[e++] = x4;                }                else                {                    q[e] = e, y[e++] = x1;                    q[e] = e, y[e++] = x4;                }            }        }        sort(r, r+c, cmpx);        int cur = 0, d = e;        double p;        for (int i = 0; i < c; i++)        {            if (!(r[i]&1))            {                if (cur == 0)                    p = x[r[i]];                cur++;            }            else            {                if (cur == 1)                {                    q[e] = e, y[e++] = p;                    q[e] = e, y[e++] = x[r[i]];                }                cur--;            }        }        sort(q, q+e, cmpy);        bool flag = 1;        int ans = cur = 0;        for (int i = 0; i < e; i++)        {            if (q[i] >= d)            {                if (!(q[i]&1))                    flag = 0;                else                    flag = 1;            }            else            {                if (!(q[i]&1))                    cur++;                else                    cur--;            }            if (flag)                ans = max(ans, cur);        }        printf("%d\n", ans);    }    return 0;}