Spoj Pouring water

Transmission Door

Pour1-pouring WATER#GCD #recursion

Given vessels, one of which can accommodate a litres of water and the other- b litres of water, det Ermine the number of steps required to obtain exactly C litres of water in one of the vessels.

At the beginning both vessels is empty. The following operations is counted as ' steps ':

• Emptying a vessel,
• Filling a vessel,
• Pouring water from one vessel to the other, without spilling, until one of the vessels are either full or empty.

Input

An integer t, 1<=t<=100, denoting the number of testcases, followed by T sets of input da TA, each consisting of three positive integers A, B, C, not larger than 40000, given in separate lines.

Output

For each set of input data, output the minimum number of steps required to obtain C litres, or-1 if this is impossible.

Examplesample Input:

2523234

Sample output:

2-1
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Solution
BFS
The most important thing to write BFS is to avoid repeating the same state of the queue.
Also check that the target status should be in each state out of the team, because the exit of the state is "unique", and the entrance generally has a number of situations (that is, the team head state can generally be transferred to a number of new States), note the bold line in the code.
In addition, because the initial state of the problem can be transferred to the state is not many (also because the two-dimensional array is not open), you should use the map to save each state to the initial state of the distance (and the minimum number of transfer steps required).
Another trick is to write Enqueue as a function that avoids the duplication of code that comes with multiple state transitions.

#include <bits/stdc++.h>using namespaceStd;typedef pair<int,int>P;intgcdintAintb) {returnB?GCD (b, a%b): A;} Map<p,int>Mp;queue<P>que;void Enque(intAintBintd) {    inttmp=mp[{a, b}]; if(!tmp| | Tmp>d) {mp[{a, B}]=D;    Que.push ({A, b}); }}BOOLOkintAintBintc) {returna==c| | b==C;}//How BFS works?voidBFsintXintYintAintBintc)    {mp.clear ();  while(!que.empty ()) Que.pop (); Mp[{x, y}]=1;    Que.push ({x, y});  while(!Que.empty ()) {P Top=que.front (); Que.pop ();intD=Mp[top]; intX=top.first, y=Top.second;  if(x==c| | Y==c) {printf ("%d\n", D-1); return ;} if(x) Enque (0, Y, d+1); if(y) enque (x,0, d+1); if(x!=a) {Enque (A, Y, D+1); if(y) {intAdd=min (Y, A-x); Enque (x+add, Y-add, d+1); }        }        if(y!=b) {Enque (x, B, D+1); if(x) {intAdd=min (x, by); Enque (x-add, Y+add, d+1); }}} puts ("-1");}intMain () {intT scanf"%d", &T);  for(intA, B, C; t--;) {scanf ("%d%d%d", &a, &b, &c); if(C>max (A, B)) {Puts ("-1");Continue;} if(c==a| | C==b) {puts ("1");Continue;} if(a==b) {Puts ("-1");Continue;} if(C%GCD (A, B)) {Puts ("-1");Continue;} BFS (0,0, A, b, c); }}

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Write this question to the main () in a few continue all write return, unexpectedly over a sample, sure enough, the sample did not hang. In the future debug should pay attention to there is no continue write return place.

Spoj Pouring water