POJ 3414 Pots (bfs+ Clues)

Pots

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

Description

You are given the pots of 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 was full (and there may be some water left in the pot I), or the PO T I was empty (and all its contents has been moved to the pot J).

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

Input

On the first and only line is the numbers A, B, and C. These is all integers in the range from 1 to + 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 is several sequences of minimal length, output any one of them. If the desired result can ' t is achieved, the first and only line of the file must contain the word 'impossible'.

Sample Input

3 5 4

Sample Output

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

Idea: A common have 6 kinds of operation: Pour a in the water, put a full, the water in B pour into a. B and A are similar.

The jar has a maximum capacity of 100, set a constant n=100, and open a two-dimensional array to record the value of the state change.

1, from the faucet to a Riga water T, record-T,

2, from the faucet to B Riga water T, record-t-n,

3, from B inside to a plus water T, record t

4, from a inside to B plus water T. Record N+t

5. Pour the water in a, record 2*n+t, (a original water t)

6, the water is poured in B, 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 105const 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 () \ 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 occur once.    The assignment is INF to 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)//to a plus water {int tmp=a-cur.x;            NEXT.Y=CUR.Y;         Next.x=a;                Water from the faucet if (!mark[next.x][next.y]) {mark[next.x][next.y]=-tmp;            Q.push (next);                } if (cur.y>0)//water from B {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)//to B plus water {int tmp=b-cur.y;            next.x=cur.x;      Next.y=b;                Water from the faucet if (!mark[next.x][next.y]) {mark[next.x][next.y]=-tmp-n;            Q.push (next);                } if (cur.x>0)//water from a {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 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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POJ 3414 Pots (bfs+ Clues)