2014 网选 5012 Dice(bfs模板)

 /*
     题意：就是给定两个筛子，每个筛子上6个面，每个面的数字属于[1,6]， 且互不相同！
     问a筛子最少经过按照题目规定的要求转动，达到和b筛子上下左右前后的数字相同！

     思路：很直白的bfs，将每一种状态对应一个数字，保证这种状态不会重新加入队列中！
 */
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<queue>
 using namespace std;

 int a[], ss;
 int vis[];

 struct node{
     int k[];
     node(){}
     node(int a1, int a2, int a3, int a4, int a5, int a6){
         k[]=a1;
         k[]=a2;
         k[]=a3;
         k[]=a4;
         k[]=a5;
         k[]=a6;
     }
     int step;
 };

 queue<node>q; 

 int sum(node x){
     int s=;
     for(int i=; i<=; ++i)
        s= s* + x.k[i];
     return s;
 }

 bool bfs(){
     while(!q.empty()) q.pop();
     node cur(a[], a[], a[], a[], a[], a[]);
     cur.step=;
     q.push(cur);
     vis[sum(cur)]=;
     while(!q.empty()){
         cur = q.front();
         if(sum(cur)==ss){
             printf("%d\n", cur.step);
             return true;
         }
         q.pop();
         node *nt = new node(cur.k[], cur.k[], cur.k[], cur.k[], cur.k[], cur.k[]);
         int v = sum(*nt);
         if(!vis[v]){
             vis[v]=;
             nt->step = cur.step + ;
             q.push(*nt);
         }

         nt = new node(cur.k[], cur.k[], cur.k[], cur.k[], cur.k[], cur.k[]);
         v = sum(*nt);
         if(!vis[v]){
             vis[v]=;
             nt->step = cur.step + ;
             q.push(*nt);
         }

         nt = new node(cur.k[], cur.k[], cur.k[], cur.k[], cur.k[], cur.k[]);
         v = sum(*nt);
         if(!vis[v]){
             vis[v]=;
             nt->step = cur.step + ;
             q.push(*nt);
         }

         nt = new node(cur.k[], cur.k[], cur.k[], cur.k[], cur.k[], cur.k[]);
         v = sum(*nt);
         if(!vis[v]){
             vis[v]=;
             nt->step = cur.step + ;
             q.push(*nt);
         }
     }
     return false;
 }

 int main(){
     while(scanf("%d%d%d%d%d%d", &a[], &a[], &a[], &a[], &a[], &a[])!=EOF){
         ss=;
         for(int i=; i<=; ++i){
             int x;
             scanf("%d", &x);
             ss = ss *  + x;
         }
         memset(vis, , sizeof(vis));
         if(!bfs())
             printf("-1\n");
     }
     return ;
 }