hdu 5012 Dice 2014 ACM/ICPC Asia Regional Xi'an Online

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BFS遍历翻转之后的状态，因为一共就654321个状态，不会TLE。

Trick 可以将六个面的数字映射成一个十进制数用来判断BFS是否被访问过，防止重复遍历。

本机debug了好久，BFS居然忘了pop()。。。

#include<iostream>#include<stdio.h>#include<cstdio>#include<stdlib.h>#include<vector>#include<string>#include<cstring>#include<cmath>#include<algorithm>#include<stack>#include<queue>#include <ctype.h>using namespace std;class dice{public:    int ary[7];    int step;   // int value;    dice()    {        memset(ary,0,sizeof(ary));        step=0;      //  value=0;    }    dice(int s)    {        memset(ary,0,sizeof(ary));        step=s;     //   value=v;    }};dice a;dice b;queue<dice> q;int ans;int vis[1000000];int done(dice d)//将每个dice上的组合map成一个整数，最大为654321，用于判断BFS是否访问过{    int num=0;    for(int i=1;i<=6;i++)    {        num=num*10+d.ary[i];    }    return num;}bool same(dice d){    for(int i=1;i<=6;i++)    {       // cout<<d.ary[i]<<" "<<b.ary[i]<<endl;        if(d.ary[i]!=b.ary[i])        {            return false;        }    }     return true;}dice rotated(int n,dice d){    dice c;    if(n==1)//left    {        int tmp=d.ary[3];        c.ary[3]=d.ary[1];        c.ary[1]=d.ary[4];        c.ary[4]=d.ary[2];        c.ary[2]=tmp;        c.ary[5]=d.ary[5];//don not forget to copy these two        c.ary[6]=d.ary[6];    }    else if(n==2)//right    {        int tmp=d.ary[2];        c.ary[1]=d.ary[3];        c.ary[4]=d.ary[1];        c.ary[2]=d.ary[4];        c.ary[3]=tmp;        c.ary[5]=d.ary[5];        c.ary[6]=d.ary[6];    }    else if(n==3)//front    {        int tmp=d.ary[5];        c.ary[5]=d.ary[1];        c.ary[1]=d.ary[6];        c.ary[6]=d.ary[2];        c.ary[2]=tmp;        c.ary[3]=d.ary[3];        c.ary[4]=d.ary[4];    }    else if(n==4)//back    {        int tmp=d.ary[5];        c.ary[6]=d.ary[1];        c.ary[2]=d.ary[6];        c.ary[5]=d.ary[2];        c.ary[1]=tmp;        c.ary[3]=d.ary[3];        c.ary[4]=d.ary[4];    }    return c;}void bfs(){    memset(vis,0,sizeof(vis));    ans=-1;    while(!q.empty()) q.pop();    a.step=0;    q.push(a);    vis[done(a)]=1;    dice tmp;    //cout<<done(a)<<endl;    while(!q.empty())    {        tmp=q.front();        q.pop();        if(same(tmp))        { //cout<<done(tmp)<<" "<<tmp.step<<endl;            ans=tmp.step;             //cout<<ans<<" ";            break;        }        for(int i=1;i<=4;i++)        {            dice d=rotated(i,tmp);            if(vis[done(d)]==0)            {                //cout<<i<<" "<<done(d)<<endl;                d.step=tmp.step+1;                q.push(d);                vis[done(d)]=1;//否则回到原先的状态会重复遍历                //cout<<d.step<<endl;            }        }    }    printf("%d\n",ans);}int main(){    freopen("input.txt","r",stdin);   // freopen("data.txt","r",stdin);   // freopen("out1.txt","w",stdout);   int t;   while(scanf("%d",&t)!=EOF)   {       a.ary[1]=t;       for(int i=2;i<=6;i++)       {           scanf("%d",&a.ary[i]);       }       for(int i=1;i<=6;i++)       {           scanf("%d",&b.ary[i]);       }       bfs();   }    return 0;}

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