Simple POJ 3414 pots,bfs+ record path.

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.

is to operate on two cups, fill, empty, pour over. The water in the two cups represents a state, then the BFS is good, and to record the path, remember each father, and then finally come back to it all over again.

The code is as follows:

#include <iostream>#include<cstring>#include<cmath>#include<queue>using namespacestd;inta,b,c;intrem[ the][ the];intshorem[10005];structstate{intb; inttype; intFaa,fab; intnum; State () {}};state sta[ the][ the];voidShowans (state *e) {    intcou=0;  while(e->a| | E->b) {Shorem[cou++]=e->type; E=&sta[e->faa][e->Fab]; } cout<<cou<<Endl;  for(inti=cou-1; i>=0;--i)Switch(Shorem[i]) { Case 1: cout<<"FILL (1) \ n";  Break;  Case 2: cout<<"FILL (2) \ n";  Break;  Case 3: cout<<"DROP (1) \ n";  Break;  Case 4: cout<<"DROP (2) \ n";  Break;  Case 5: cout<<"pour () \ n";  Break;  Case 6: cout<<"pour (2,1) \ n";  Break; }}voidSlove () {Queue<state *>que; State*temp; intT1,t2,t3; sta[0][0].faa=sta[0][0].fab=-1; sta[0][0].type=0; sta[0][0].num=0; Que.push (&sta[0][0]);  while(!Que.empty ()) {Temp=Que.front ();        Que.pop (); if(temp->a==c| | temp->b==C) {Showans (temp); return; } T1=temp->A; T2=temp->b; if(sta[a][t2].num==-1) {Sta[a][t2].num=temp->num+1; STA[A][T2].FAA=T1; Sta[a][t2].fab=T2; Sta[a][t2].type=1; Que.push (&Sta[a][t2]); }        if(sta[t1][b].num==-1) {Sta[t1][b].num=temp->num+1; STA[T1][B].FAA=T1; Sta[t1][b].fab=T2; Sta[t1][b].type=2; Que.push (&Sta[t1][b]); }        if(sta[0][t2].num==-1) {sta[0][t2].num=temp->num+1; sta[0][t2].faa=T1; sta[0][t2].fab=T2; sta[0][t2].type=3; Que.push (&sta[0][t2]); }        if(sta[t1][0].num==-1) {sta[t1][0].num=temp->num+1; sta[t1][0].faa=T1; sta[t1][0].fab=T2; sta[t1][0].type=4; Que.push (&sta[t1][0]); } T3=min (t1,b-T2); if(sta[t1-t3][t2+t3].num==-1) {sta[t1-t3][t2+t3].num=temp->num+1; Sta[t1-t3][t2+t3].faa=T1; Sta[t1-t3][t2+t3].fab=T2; Sta[t1-t3][t2+t3].type=5; Que.push (&sta[t1-t3][t2+T3]); } T3=min (t2,a-t1); if(sta[t1+t3][t2-t3].num==-1) {sta[t1+t3][t2-t3].num=temp->num+1; Sta[t1+t3][t2-t3].faa=T1; Sta[t1+t3][t2-t3].fab=T2; Sta[t1+t3][t2-t3].type=6; Que.push (&sta[t1+t3][t2-T3]); }} cout<<"impossible\n";}intMain () {Ios::sync_with_stdio (false);  for(intI=0; i<= -;++i) for(intj=0; j<= -;++j) {sta[i][j].a=i; STA[I][J].B=J; }     while(cin>>a>>b>>C) { for(intI=0; i<= -;++i) for(intj=0; j<= -;++j) Sta[i][j].num=-1;    Slove (); }    return 0;}

View Code

Simple POJ 3414 pots,bfs+ record path.