poj 3414 Pots （ bfs ）

题目：http://poj.org/problem?id=3414

题意：给出了两个瓶子的容量A,B, 以及一个目标水量C，

对A、B可以有如下操作:

FILL(i)        fill the pot i (1 ≤ i ≤ 2) from the tap;

DROP(i)      empty the pot i to the drain;

POUR(i,j)    pour from pot i to pot j; after this operation either the pot j is full (and there may be some water left in the pot i), or the pot i is empty (and all its contents have been moved to the pot j).

问经过哪几个操作后能使得任意一个瓶子的残余水量为C。

若不可能得到则输出impossible

 #include <iostream>
  #include<cstdio>
  #include<cstring>
  #include<cstdlib>
  #include<stack>
  #include<queue>
  #include<iomanip>
  #include<cmath>
  #include<map>
  #include<vector>
  #include<algorithm>
  using namespace std;

  int vis[][];
  int a,b,c;
  struct node
  {
      int x,y,step;
  }pos,next;

  struct way
  {
      int x,y,f;
  }before[][];

  void pri(int x,int y)
  {
      stack<int>st;
      while(x!=||y!=)
      {
          st.push(before[x][y].f);
          int  tx=before[x][y].x;
          int ty=before[x][y].y;
          x=tx; y=ty;
      }
      while(!st.empty())
      {
          switch(st.top())
          {
              case :printf("DROP(1)\n"); break;
              case :printf("DROP(2)\n"); break;
              case :printf("FILL(1)\n"); break;
              case :printf("FILL(2)\n"); break;
              case :printf("POUR(1,2)\n"); break;
              case :printf("POUR(2,1)\n"); break;
          }
          st.pop();
      }
  }
  int bfs()
  {
      queue<node>q;
      next.x=; next.y=; next.step=;
      vis[][]=;
      q.push(next);
      while(!q.empty())
      {
          pos=q.front();
          //cout<<pos.x<<" "<<pos.y<<" "<<pos.step<<"    ";
          //cout<<before[pos.x][pos.y].x<<" "<<before[pos.x][pos.y].y<<" "<<before[pos.x][pos.y].f<<endl;
          q.pop();
          if(pos.x==c||pos.y==c)
          {
              cout<<pos.step<<endl;
              pri(pos.x,pos.y);
              return ;
          }
          if(!vis[][pos.y]&&pos.x!=)
          {
              next.x=; next.y=pos.y; next.step=pos.step+;
              q.push(next);
              vis[][pos.y]=;
              before[][pos.y]=(struct way){pos.x,pos.y,};
          }
          if(!vis[pos.x][]&&pos.y!=)
          {
              next.x=pos.x; next.y=; next.step=pos.step+;
              q.push(next);
              vis[pos.x][]=;
              before[pos.x][]=(struct way){pos.x,pos.y,};
          }
          if(!vis[a][pos.y]&&pos.x!=a)
          {
              next.x=a; next.y=pos.y; next.step=pos.step+;
              q.push(next);
              vis[a][pos.y]=;
              before[a][pos.y]=(struct way){pos.x,pos.y,};
          }
          if(!vis[pos.x][b]&&pos.y!=b)
          {
              next.x=pos.x; next.y=b; next.step=pos.step+;
              q.push(next);
              vis[pos.x][b]=;
              before[pos.x][b]=(struct way){pos.x,pos.y,};
          }
          if(pos.x>&&pos.y<b)
          {
              int t=min(pos.x,b-pos.y);
              if(!vis[pos.x-t][pos.y+t])
              {
                  q.push((struct node){pos.x-t,pos.y+t,pos.step+});
                  vis[pos.x-t][pos.y+t]=;
                  before[pos.x-t][pos.y+t]=(struct way){pos.x,pos.y,};
              }
          }
          if(pos.x<a&&pos.y>)
          {
              int t=min(pos.y,a-pos.x);
              if(!vis[pos.x+t][pos.y-t])
              {
                  q.push((struct node){pos.x+t,pos.y-t,pos.step+});
                  vis[pos.x+t][pos.y-t]=;
                  before[pos.x+t][pos.y-t]=(struct way){pos.x,pos.y,};
              }
          }
      }
      return ;
  }
  int main()
  {
      cin>>a>>b>>c;
      memset(vis,,sizeof(vis));
      if(bfs()==)
      cout<<"impossible"<<endl;
      return ;
  }