HDU 4445 Crazy Tank 三分+枚举+二分

题意：在高为H的台上有一个坦克向右打炮，[L1.R1]为敌军范围，[L2,R2]为友军范围，现在有n（0<=n<=200）个初速度不同的炮弹，现在想找到一个最好的

         角度使得任意一个不打在友军的范围内同时打在敌军最多。

题解：对于不同的炮弹打的最远的角度是不一样的，因为这个我wa的很惨。对于每个炮弹三分找出能打最远的角度，然后枚举打的点分别为L1.R1,L2,R2，

         二分得到两个角度更新答案。

         PS：个人认为出题者的本意不应该是暴力枚举角度吧……

Sure原创，转载请注明出处

#include <iostream>#include <cstdio>#include <memory.h>#include <algorithm>#include <cmath>#define MAX(a , b) ((a) > (b) ? (a) : (b))using namespace std;const int maxn = 202;const double g = 9.80;const double eps = 1e-8;const double PI = acos(-1.0);double pos[10],v[maxn];double h,farest;int n;void read(){    scanf("%lf %lf %lf %lf %lf",&h,&pos[0],&pos[1],&pos[2],&pos[3]);    for(int i=0;i<n;i++)    {        scanf("%lf",&v[i]);    }    return;}double dis(double ang,double vv){    double s = sin(ang);    double c = cos(ang);    return vv * s * (vv * c + sqrt(vv * vv * c * c + 2.0 * g * h)) / g;}void tris(double vv){    double l = 0,r = PI / 2.0;    int t = 30;    while(t--)    {        double mid = (l + r) / 2.0;        double midd = (l + mid) / 2.0;        if(dis(midd , vv) > dis(mid , vv)) r = mid;        else l = midd;    }    farest = l;    return;}double bis_left(double l,double r,double d,double vv){    int t = 30;    while(t--)    {        double mid = (l + r) / 2.0;        if(dis(mid , vv) <= d) l = mid;        else r = mid;    }    return l;}double bis_right(double l,double r,double d,double vv){    int t = 30;    while(t--)    {        double mid = (l + r) / 2.0;        if(dis(mid , vv) < d) r = mid;        else l = mid;    }    return l;}void solve(){    int res = 0;    for(int i=0;i<n;i++)    {        for(int j=0;j<4;j++)        {            tris(v[i]);            if(dis(farest , v[i]) < pos[j]) continue;            double wei = pos[j];            if(j == 0) wei += eps;            if(j == 1) wei -= eps;            if(j == 2) wei -= eps;            if(j == 3) wei += eps;            double ang = bis_left(0.0 , farest , wei , v[i]);            bool flag = true;            int c = 0;            for(int k=0;k<n;k++)            {                double here = dis(ang , v[k]);                if(here >= pos[2] && here <= pos[3])                {                    flag = false;                    break;                }                else if(here >= pos[0] && here <= pos[1]) c++;            }            if(flag) res = MAX(res , c);            if(res == n) break;            ang = bis_right(farest , PI , wei , v[i]);            flag = true;            c = 0;            for(int k=0;k<n;k++)            {                double here = dis(ang , v[k]);                if(here >= pos[2] && here <= pos[3])                {                    flag = false;                    break;                }                else if(here >= pos[0] && here <= pos[1]) c++;            }            if(flag) res = MAX(res , c);            if(res == n) break;        }    }    printf("%d\n",res);    return;}int main(){    while(scanf("%d",&n) && n)    {        read();        solve();    }    return 0;}