HDU 4458 Shoot the Airplane （2012年杭州赛区现场赛F题）

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1.题目描述：点击打开链接

2.解题思路：本题利用运动的相对性解决。可以假设飞机固定不动，那么子弹相当于还有一个水平分量，只不过方向是-v。这样，问题就转化为一个动点什么时候到达一个多边形内部的问题了。不过本题有一个细节就是g可能等于0，因此要分匀变速运动和匀速运动2种情况求出最大可能的子弹飞行时间。另外一个细节就是本题对精度要求比较高，建议判断点是否在线段上改为利用坐标的差值来判断，避开使用Dot函数。

3.代码：

#include<iostream>#include<algorithm>#include<cassert>#include<string>#include<sstream>#include<set>#include<bitset>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<cctype>#include<functional>#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define me(s)  memset(s,0,sizeof(s))#define rep(i,n) for(int i=0;i<(n);i++)typedef long long ll;typedef unsigned int uint;typedef unsigned long long ull;//typedef pair <int, int> P;const double PI=acos(-1.0);const double eps=1e-10;int dcmp(double x){    if(fabs(x)<eps)return 0;    return x<0?-1:1;}struct Point{    double x,y;    Point(){}    Point(double x,double y):x(x),y(y){}    Point operator+(const Point&p){return Point(x+p.x,y+p.y);}    Point operator-(const Point&p){return Point(x-p.x,y-p.y);}    Point operator*(double p){return Point(x*p,y*p);}    Point operator/(double p){return Point(x/p,y/p);}};typedef Point Vector;double Dot(Vector a,Vector b){return a.x*b.x+a.y*b.y;}double Cross(Vector a,Vector b){return a.x*b.y-a.y*b.x;}double Length(Vector a){return sqrt(Dot(a,a));}double Angle(Vector a,Vector b){return acos(Dot(a,b)/Length(a)/Length(b));}bool OnSegment(Point p,Point a,Point b){    if(dcmp(Cross(p-a,p-b)))return 0;    return dcmp(a.x-p.x)*dcmp(b.x-p.x)<=0&&dcmp(a.y-p.y)*dcmp(b.y-p.y)<=0; //利用dcmp判断，避免使用Dot}typedef vector<Point> Polygon;int isPointInPolygon(Point p,Polygon poly){    int wn=0;    int n=poly.size();    for(int i=0;i<n;i++)    {        if(OnSegment(p,poly[i],poly[(i+1)%n]))return 0;        int k=dcmp(Cross(poly[(i+1)%n]-poly[i],p-poly[i]));        int d1=dcmp(poly[i].y-p.y);        int d2=dcmp(poly[(i+1)%n].y-p.y);        if(k>0&&d1<=0&&d2>0)wn++;        if(k<0&&d2<=0&&d1>0)wn--;    }    return wn;}double v,b,g;int n;int main(){    while(~scanf("%lf%lf%lf",&v,&b,&g))    {        if(v==0&&b==0&&g==0)break;        scanf("%d",&n);        Polygon p;        double my=0.0,x,y;        for(int i=0;i<n;i++)        {            scanf("%lf%lf",&x,&y);            my=max(my,y);            p.push_back(Point(x,y));        }        int ok=0;        double T=dcmp(g)?2.0*b/g:my/b; //分2种情况        for(double t=0.0;t<=T;t+=0.001)        {            Point tmp(-v*t,b*t-0.5*g*t*t);             if(isPointInPolygon(tmp,p))            {                printf("%.2lf\n",t);                ok=1;                break;            }        }        if(!ok)puts("Miss!");    }}

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