POJ 3414 Pots

Pots

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12022		Accepted: 5078		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 potJ was full (and there may be some water left in the pot I), or the pot 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 would yield exactlyC liters of W Ater in one of the pots.

Input

On the first and only line is the numbers A, B, andC. 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 operationsK. 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)

The main topic: there are two cups of the capacity of a, B, ask how many steps, you can let C-liter of water into a cup.

Idea: Ask the shortest steps of course is BFS, but how to mark down its path is difficult, and how a search method.

You can set a string in node to specifically install the current situation, and then map to STR, for how the search is definitely divided into fill,drop,pour three categories

#include <iostream> #include <cstdio> #include <string.h> #include <string> #include <cmath    > #include <queue> #define LL long longusing namespace std;int a,b,c;struct node{int n1,n2,tep; String a;//saves the current operation};bool Vis[100][100];char str[6][10]= {"Fill (1)", "Fill (2)", "pour (2,1)", "pour", "DROP (1)", "    DROP (2) "};//specifically has six implementations of void BFs () {queue<node>q;    while (!q.empty ()) Q.pop ();    Node F1,f2;    int t;    f1.n1=0;    f1.n2=0;    F1.tep=0;    F1.a= ""; Start with an empty string Q.push (F1);        while (!q.empty ()) {F1=q.front ();        Q.pop ();       if (VIS[F1.N1][F1.N2])//paper-cut continue;        else vis[f1.n1][f1.n2]=true; if (f1.n1==c| |            F1.n2==c) {printf ("%d\n", f1.tep);            for (int j=0;j<f1.tep;j++) {printf ("%s\n", str[f1.a[j]-' 0 ");//corresponding to STR}        return;            } for (int i=0;i<4;i++) {f2=f1;            f2.tep++; if (i==0)           {if (f2.n1==0) {f2.n1=a;                    f2.n2=f1.n2;//Note that this must be f2.x=f1.x because it is not possible to affect the next if (), although the beginning of the part of the F2=F1, in the bottom side of the section is not the f2.a=f1.a+ ' 0 ';                Q.push (F2);                    } if (f2.n2==0) {f2.n2=b;                    f2.n1=f1.n1;//remains relatively constant f2.a=f1.a+ ' 1 ';                Q.push (F2);                    }} if (I==1) {if (f2.n1!=0) {f2.n1=0;                    F2.N2=F1.N2;                    f2.a=f1.a+ ' 4 ';                Q.push (F2);                    } if (f2.n2!=0) {f2.n2=0;                    F2.N1=F1.N1;                    f2.a=f1.a+ ' 5 ';                Q.push (F2);          }} if (i==3) {if (f2.n2!=0&&f2.n1!=a) {          T=A-F2.N1;                        if (f2.n2>=t) {f2.n2-=t;                    F2.n1=a;                        } else {f2.n1+=f2.n2;                    f2.n2=0;                    } f2.a+= ' 2 ';                Q.push (F2);                    }} if (i==2) {if (f2.n1!=0&&f2.n2!=b) {                    T=B-F2.N2;                        if (f2.n1>=t) {f2.n1-=t;                    F2.n2=b;                        } else {f2.n2+=f2.n1;                    f2.n1=0;                    } f2.a+= ' 3 ';                Q.push (F2); }}}} printf ("impossible\n");}    int main () {int n,m,i,j,k,s; while (~SCANF ("%d%d%d",&a,&b,&C) {memset (vis,false,sizeof (VIS));    BFS (); } return 0;}

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POJ 3414 Pots