POJ 3414 Pots (bfs+ recursive printing)

Pots

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10255		Accepted: 4333		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)

Test instructions: Give you two known volumes of cups A, B, and a C, for a minimum number of steps to derive c capacity of water. Fill (i) Fill I DROP (i): pour light I pour (i,j): pour i to J
The following: BFS, a total of 6 cases: Fill (a), fill (b), drop (a), drop (b), pour (a), pour (b,a). Among them, pour (i,j) is divided into two kinds of situation and pour full.

#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <string > #include <algorithm> #include <cstdlib> #include <set> #include <queue> #include <stack > #include <vector> #include <map> #define N 100010#define Mod 10000007#define Lson l,mid,idx<<1#  Define Rson mid+1,r,idx<<1|1#define LC idx<<1#define RC Idx<<1|1const Double EPS = 1e-11;const double PI = ACOs ( -1.0); const double E = 2.718281828;typedef long long ll;const int INF = 1000010;using namespace Std;int n,m,x;s Tring s[3]= {"Fill", "DROP", "pour"};bool vis[110][110];struct node{int num,nex;///num==:1 (fill), 2 (Drop), 3 (pour) int t;///full, guang bottle int x1,x2;///num=3, by x1->x2 int a,b;///a,b State int step;///Steps} mp[1010];queue<node>que;in    T Fro,nex;bool BFs () {while (Que.size ()) Que.pop ();    memset (vis,0,sizeof Vis);    Node A;    Fro=0,nex=1; MP[FRO]. A=0,MP[FRO].    b=0;    Mp[fro].nex=-1; Mp[fro].step=0;    Que.push (Mp[fro]);    Vis[0][0]=1;        while (Que.size ()) {A=que.front ();        Que.pop (); if (a.a==x| |            a.b==x) {printf ("%d\n", a.step);        return 1;        } mp[nex].nex=-1;            if (!vis[n][a.b])///Pour full a {mp[nex].nex=fro; Mp[nex]. A=n,mp[nex].            B=A.B;            Mp[nex].num=1;            Mp[nex].t=1;            mp[nex].step=a.step+1;            Que.push (Mp[nex]);            Vis[n][a.b]=1;        nex++;            } if (!vis[a.a][m])///pour full B {mp[nex].nex=fro; Mp[nex]. A=a.a,mp[nex].            B=m;            Mp[nex].num=1;            mp[nex].t=2;            mp[nex].step=a.step+1;            Que.push (Mp[nex]);            Vis[a.a][m]=1;        nex++;            } if (!vis[0][a.b])///pour light a {mp[nex].nex=fro; Mp[nex]. A=0,mp[nex].            B=A.B;            mp[nex].num=2;            Mp[nex].t=1;            mp[nex].step=a.step+1;           Que.push (Mp[nex]); Vis[0][a.b]=1;        nex++;            } if (!vis[a.a][0])///pour light B {mp[nex].nex=fro; Mp[nex]. A=a.a,mp[nex].            b=0;            mp[nex].num=2;            mp[nex].t=2;            mp[nex].step=a.step+1;            Que.push (Mp[nex]);            Vis[a.a][m]=1;        nex++; } if (a.a>0)///a->b {int t=m-a.            B if (a.a>t)//{Mp[nex].                A=a.a-t; Mp[nex].            B=m; } else//{Mp[nex].                a=0; Mp[nex].            B=A.A+A.B; } if (!vis[mp[nex]. A][mp[nex].                B]) {mp[nex].num=3;                mp[nex].x1=1,mp[nex].x2=2;                Mp[nex].nex=fro;                mp[nex].step=a.step+1;                Que.push (Mp[nex]); Vis[mp[nex]. A][mp[nex].                B]=1;            nex++; }} if (a.b>0)///b->a {int t=n-a.            A if (a.b&GT;T)//{Mp[nex].                A=n; Mp[nex].            B=a.b-t; } else//{Mp[nex].                A=A.A+A.B; Mp[nex].            b=0; } if (!vis[mp[nex]. A][mp[nex].                B]) {mp[nex].num=3;                Mp[nex].x1=2,mp[nex].x2=1;                Mp[nex].nex=fro;                mp[nex].step=a.step+1;                Que.push (Mp[nex]); Vis[mp[nex]. A][mp[nex].                B]=1;            nex++;    }} fro++; } return 0;}        void print (int x) {if (mp[x].nex!=-1) {print (Mp[x].nex);        if (mp[x].num==1) printf ("FILL (%d) \ n", mp[x].t);        else if (mp[x].num==2) printf ("DROP (%d) \ n", mp[x].t);    else printf ("pour (%d,%d) \ n", mp[x].x1,mp[x].x2); }}int Main () {while (cin>>n>>m>>x) {if (!bfs ()) cout<< "Impossible" <<e        Ndl else {print (fro);       }} return 0;} 

POJ 3414 Pots (bfs+ recursive printing)