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[110][110];
int a,b,c;
struct node
{
int x,y,step;
}pos,next;

struct way
{
int x,y,f;
}before[110][110];

void pri(int x,int y)
{
stack<int>st;
while(x!=0||y!=0)
{
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 1:printf("DROP(1)\n"); break;
case 2:printf("DROP(2)\n"); break;
case 3:printf("FILL(1)\n"); break;
case 4:printf("FILL(2)\n"); break;
case 5:printf("POUR(1,2)\n"); break;
case 6:printf("POUR(2,1)\n"); break;
}
st.pop();
}
}
int bfs()
{
queue<node>q;
next.x=0; next.y=0; next.step=0;
vis[0][0]=1;
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 1;
}
if(!vis[0][pos.y]&&pos.x!=0)
{
next.x=0; next.y=pos.y; next.step=pos.step+1;
q.push(next);
vis[0][pos.y]=1;
before[0][pos.y]=(struct way){pos.x,pos.y,1};
}
if(!vis[pos.x][0]&&pos.y!=0)
{
next.x=pos.x; next.y=0; next.step=pos.step+1;
q.push(next);
vis[pos.x][0]=1;
before[pos.x][0]=(struct way){pos.x,pos.y,2};
}
if(!vis[a][pos.y]&&pos.x!=a)
{
next.x=a; next.y=pos.y; next.step=pos.step+1;
q.push(next);
vis[a][pos.y]=1;
before[a][pos.y]=(struct way){pos.x,pos.y,3};
}
if(!vis[pos.x][b]&&pos.y!=b)
{
next.x=pos.x; next.y=b; next.step=pos.step+1;
q.push(next);
vis[pos.x][b]=1;
before[pos.x][b]=(struct way){pos.x,pos.y,4};
}
if(pos.x>0&&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+1});
vis[pos.x-t][pos.y+t]=1;
before[pos.x-t][pos.y+t]=(struct way){pos.x,pos.y,5};
}
}
if(pos.x<a&&pos.y>0)
{
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+1});
vis[pos.x+t][pos.y-t]=1;
before[pos.x+t][pos.y-t]=(struct way){pos.x,pos.y,6};
}
}
}
return 0;
}
int main()
{
cin>>a>>b>>c;
memset(vis,0,sizeof(vis));
if(bfs()==0)
cout<<"impossible"<<endl;
return 0;
}