bzoj2144: 跳跳棋

昨天考试的神仙题

对于一个状态(x,y,z)，有三种转移方案，往外跳两种，往里跳只有1种(考试的时候没有意识到)

那么可以看作一棵树，往外跳是子节点，往里跳是父亲

问题转换成树上两个点求最短路，这样就只用往里面跳了

考虑往里面跳是相当于一个辗转相除的，复杂度是logK，根据求LCA倍增的思想，不停往上跳就好。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
bool check(int ax,int ay,int az,int bx,int by,int bz)
{
    int t1,t2,w;
    
    t1=ay-ax,t2=az-ay,w=0;
    if(t1>t2){swap(t1,t2);w^=1;}
    
    while((t2-1)/t1>0)
    {
        int c=(t2-1)/t1;
        
        if(w==0){ax+=c*t1;ay+=c*t1;}
        else     {ay-=c*t1,az-=c*t1;}
        
        t2=t2-t1*c;
        swap(t1,t2);w^=1;
    }
    
    t1=by-bx,t2=bz-by,w=0;
    if(t1>t2){swap(t1,t2);w^=1;}
    
    while((t2-1)/t1>0)
    {
        int c=(t2-1)/t1;
        
        if(w==0){bx+=c*t1;by+=c*t1;}
        else     {by-=c*t1;bz-=c*t1;}
        
        t2=t2-t1*c;
        swap(t1,t2);w^=1;
    }
    
    return ax==bx&&ay==by&&az==bz;
}

int getdep(int x,int y,int z)
{
    int t1=y-x,t2=z-y,ret=0;
    if(t1>t2)swap(t1,t2);
    while((t2-1)/t1>0)
    {
        int c=(t2-1)/t1;
        ret+=c;
        t2=t2-t1*c;
        swap(t1,t2);
    }
    return ret;
}
void jumpK(int &x,int &y,int &z,int d)
{
    int t1=y-x,t2=z-y,w=0;
    if(t1>t2){swap(t1,t2);w^=1;}
    while(d>0)
    {
        int c=min(d,(t2-1)/t1);
        d-=c;
        
        if(w==0){x+=c*t1;y+=c*t1;}
        else     {y-=c*t1,z-=c*t1;}
        
        t2=t2-t1*c;
        swap(t1,t2);w^=1;
    }
}
int main()
{
    freopen("2.in","r",stdin);
    freopen("2.out","w",stdout);
    int ax,ay,az,bx,by,bz;
    scanf("%d%d%d%d%d%d",&ax,&ay,&az,&bx,&by,&bz);
    if(ax>ay)swap(ax,ay);                     if(bx>by)swap(bx,by);
    if(ax<az&&az<ay)swap(ay,az);              if(bx<bz&&bz<by)swap(by,bz);
    if(az<ax)swap(ay,az),swap(ax,ay);         if(bz<bx)swap(by,bz),swap(bx,by);
    
    if(!check(ax,ay,az,bx,by,bz)){printf("NO\n");return 0;}
    else printf("YES\n");
    
    //--------------------------------------
    
    int da=getdep(ax,ay,az),db=getdep(bx,by,bz);
    int ans=0;
    if(da!=db)
    {
        if(da<db)swap(da,db),swap(ax,bx),swap(ay,by),swap(az,bz);
        ans+=da-db;
        jumpK(ax,ay,az,da-db);
    }
    if(ax==bx&&ay==by&&az==bz){printf("%d\n",ans);return 0;}
    for(int i=30;i>=0;i--)
        if(db>(1<<i))
        {
            int tax=ax,tay=ay,taz=az;
            int tbx=bx,tby=by,tbz=bz;
            jumpK(ax,ay,az,(1<<i));
            jumpK(bx,by,bz,(1<<i));
            if(ax==bx&&ay==by&&az==bz)
            {
                ax=tax,ay=tay,az=taz;
                bx=tbx,by=tby,bz=tbz;
            }
            else ans+=(1<<i)*2,db-=(1<<i);
        }
    printf("%d\n",ans+2);
    return 0;
}