POJ 3414 pots (bfs+ clues)

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Pots

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10071		Accepted: 4237		Special Judge

Description

You are given two pots, having the volume of A and B liters respectively. The following operations can be performed:

1. FILL(i) fill the pot i (1 ≤ i ≤ 2) from the tap;
2. DROP(i) empty the pot i to the drain;
3. 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).

Write a program to find the shortest possible sequence of these operations that will yield exactly C liters of water in one of the pots.

Input

On the first and only line are the numbers A, B, and C. These are all integers in the range from 1 to 100 and C≤max(A,B).

Output

The first line of the output must contain the length of the sequence of operations K. The following K lines must each describe one operation. If there are several sequences of minimal length, output any one of them. If the desired result can’t be achieved, the first and only line of the file must contain the word ‘impossible’.

Sample Input

3 5 4

Copy

Sample Output

6
FILL(2)
POUR(2,1)
DROP(1)
POUR(2,1)
FILL(2)
POUR(2,1)

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Ideas ： We all have 6 Kind of operation ： hold A The water is poured out , hold A Fill it up with , hold B Pour the water in A in .B and A similar .

The maximum volume of the tank is 100, Set a constant N=100, Open a two-dimensional array to record the value of state change .

1、 From the tap to A Add water in t, Record -t,

2、 From the tap to B Add water in t, Record -t-N,

3、 from B Inside A Add water t, Record t

4、 from A Inside B Add water t. Record N+t

5、 hold A Pour out the water , Record 2*N+t,(A Original water t)

6、 hold B Pour out the water , Record 3*N+t,(B Original water t)

#include<stdio.h>
#include<queue>
#include<map>
#include<string>
#include<string.h>
using namespace std;
#define N 105
const int inf=0x1f1f1f1f;
int a,b,c,flag;
int mark[N][N];
struct node
{
    int x,y,t;
    friend bool operator<(node a,node b)
    {
        return a.t>b.t;
    }
};
void prif(int x,int y)       // Recursive output path 
{
    if(x==0&&y==0)
        return ;
    if(mark[x][y]>3*N)
    {
        prif(x,mark[x][y]-3*N);
        printf("DROP(2)\n");
    }
    else if(mark[x][y]>2*N)
    {
        prif(mark[x][y]-2*N,y);
        printf("DROP(1)\n");
    }
    else if(mark[x][y]>N)
    {
        int tmp=mark[x][y]-N;
        prif(x+tmp,y-tmp);
        printf("POUR(1,2)\n");
    }
    else if(mark[x][y]>0)
    {
        int tmp=mark[x][y];
        prif(x-tmp,y+tmp);
        printf("POUR(2,1)\n");
    }
    else if(mark[x][y]>-N)
    {
        int tmp=-mark[x][y];
        prif(x-tmp,y);
        printf("FILL(1)\n");
    }
    else if(mark[x][y]<-N)
    {
        int tmp=N+mark[x][y];
        prif(x,y+tmp);
        printf("FILL(2)\n");
    }
}
void bfs()
{
    priority_queue<node>q;
    node cur,next;
    mark[0][0]=inf;  // This state can only appear once . The assignment is inf Prevent interference with other values 
    mark[a][0]=-a;
    mark[0][b]=-b-N;
    cur.t=1;       
    cur.x=a;
    cur.y=0;
    q.push(cur);
    cur.x=0;
    cur.y=b;
    q.push(cur);
    while(!q.empty())
    {
        cur=q.top();
        q.pop();
        next.t=cur.t+1;
        if(cur.x==c||cur.y==c)
        {
            flag=1;
            printf("%d\n",cur.t);
            prif(cur.x,cur.y);
            return ;
        }
        if(cur.x<a)       // towards A Add water 
        {
            int tmp=a-cur.x;
            next.y=cur.y;
            next.x=a;         // Water from the tap 
            if(!mark[next.x][next.y])
            {
                mark[next.x][next.y]=-tmp;
                q.push(next);
            }
            if(cur.y>0)      // come from B Of water 
            {
                int tmp=min(cur.y,a-cur.x);
                next.x=cur.x+tmp;
                next.y=cur.y-tmp;
                if(!mark[next.x][next.y])
                {
                    mark[next.x][next.y]=tmp;
                    q.push(next);
                }
            }
        }
        if(cur.y<b)        // towards B Add water 
        {
            int tmp=b-cur.y;
            next.x=cur.x;
            next.y=b;      // Water from the tap 
            if(!mark[next.x][next.y])
            {
                mark[next.x][next.y]=-tmp-N;
                q.push(next);
            }
            if(cur.x>0)      // come from A Of water 
            {
                int tmp=min(cur.x,b-cur.y);
                next.y=cur.y+tmp;
                next.x=cur.x-tmp;
                if(!mark[next.x][next.y])
                {
                    mark[next.x][next.y]=tmp+N;
                    q.push(next);
                }
            }
        }
        if(cur.x>0)     // Pour out the water 
        {
            int tmp=cur.x;
            next.x=0;
            next.y=cur.y;
            if(!mark[next.x][next.y])
            {
                mark[next.x][next.y]=2*N+tmp;
                q.push(next);
            }
        }
        if(cur.y>0)
        {
            int tmp=cur.y;
            next.y=0;
            next.x=cur.x;
            if(!mark[next.x][next.y])
            {
                mark[next.x][next.y]=3*N+tmp;
                q.push(next);
            }
        }
    }
}
int main()
{
    while(scanf("%d%d%d",&a,&b,&c)!=-1)
    {
        memset(mark,0,sizeof(mark));
        flag=0;
        bfs();
        if(!flag)
            printf("impossible\n");
    }
    return 0;
}

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