【结论+简单几何】HDU6147 Pokémon GO II

题面在这里

不难发现，第一次路径相交时，这两条线段出现的时间差不超过6

然后就很好搞了

注意判断线段相交有很多小细节

示例程序：

#include<cstdio>#include<cstring>#include<algorithm>#define LL long longusing namespace std;inline char nc(){    static char buf[100000],*p1=buf,*p2=buf;    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline int red(){    int res=0,f=1;char ch=nc();    while (ch<'0'||'9'<ch) {if (ch=='-') f=-f;ch=nc();}    while ('0'<=ch&&ch<='9') res=res*10+ch-48,ch=nc();    return res*f;}const int p[4][2]={{-1,0},{0,1},{1,0},{0,-1}};int tst,n;struct seg{    LL x,y,xx,yy;    seg () {}    seg (LL _x,LL _y,LL _xx,LL _yy):x(_x),y(_y),xx(_xx),yy(_yy) {}    void maintain(){        if (x==xx){            if (y>yy) swap(y,yy);        }else if (x>xx) swap(x,xx);    }    bool check(const seg&b){        if (x==xx&&y==yy) return 0;        if (x==xx){            if (b.y==b.yy) return y<=b.y&&b.y<=yy && b.x<=x&&x<=b.xx;             else return x==b.x&&!(b.yy<y||yy<b.y);        }else{            if (b.x==b.xx) return x<=b.x&&b.x<=xx && b.y<=y&&y<=b.yy;             else return y==b.y&&!(b.xx<x||xx<b.x);        }    }}a[10];int main(){    tst=red();    while (tst--){        n=red();memset(a,0,sizeof(a));        int til=0,i=0;LL x=0,y=0;        bool suc=1;        for (int j=1;j<=n;j++){            LL l=red();            a[til]=seg(x,y,x+p[i][0]*l,y+p[i][1]*l);            a[til].maintain();            x+=p[i][0]*l;y+=p[i][1]*l;i=(i+1)%4;            for (int k=0;k<6;k++)             if (k!=til&&k!=(til+5)%6&&a[k].check(a[til])) {printf("%d\n",j);suc=0;break;}            if (!suc) while (++j<=n) red();            til=(til+1)%6;        }        if (suc) printf("Catch you\n");    }    return 0;}