Milk pails (BFS)

Milk pails

Time limit: 1 Sec memory limit: up to MB
Submitted: 4 Resolution:
Submitted State [Discussion Version]

Title DescriptionFarmer John has received a order for exactly M units of milk (1≤m≤200) that he needs to fill right away. Unfortunately, his fancy milking machine had just become broken, and all he had is is both milk pails of integer sizes X and Y (1≤x,y≤100) with which he can measure milk. Both Pails is initially empty. Using These-pails, he can perform up to K of the following types of operations (1≤K≤100):

-He can fill either pail completely to the top.

-He can empty either pail.

-He can pour the contents of one pail into the other, stopping when the former becomes empty or the latter becomes full ( Whichever of these happens first).

Although FJ realizes he is not being able to end up with exactly M total units of milk in the both pails, please help him Compute the minimum amount of error between M and the total amount of milk in the pails. That's, please compute the minimum value of | m−m′| such that FJ can construct m′units of milk collectively between the pails.
InputThe first, and only line of input, contains X, Y, K, and M.
OutputOutput the smallest distance from M to an amount of milk FJ can produce.
Sample input

14 50 2 32

Sample output

18
"Analysis" This is particularly similar to yesterday's mother's milk, two bottles, initially empty, there are three kinds of operations, a bottle filled with a bottle empty, and then will be a bottle net another bottle pour, until the end
or to full. Changed yesterday's code a little bit.

#include <iostream>#include<cstring>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<time.h>#include<string>#include<map>#include<stack>#include<vector>#include<Set>#include<queue>#defineINF 0x3f3f3f3f#defineMoD 1000000007typedefLong Longll;using namespacestd;Const intn= About;intN,dp[n],len;intW[n][n];intg[3];inta,b,c;intx,y,k,m;Set<int>p;intminn=inf;structMan {intx, y; intstep;};voidBFs () {w[0][0]=1; Queue<man>Q;    Man S; S.x=0; S.y=0; S.step=0;    Q.push (s);  while(!Q.empty ()) {Man T=Q.front ();        Q.pop (); Minn=min (Minn,abs (T.X+T.Y)-M)); if(t.step==k)Continue; intf[3]; f[0]=T.x; f[1]=T.y; if(!w[f[0]][0]) {man kk;kk.x=f[0];kk.y=0; kk.step=t.step+1; w[f[0]][0]=1; Q.push (KK);} if(!w[f[0]][g[1]]) {man kk;kk.x=f[0];kk.y=g[1];kk.step=t.step+1; w[f[0]][g[1]]=1; Q.push (KK);} if(!w[0][f[1]]) {man kk;kk.x=0; kk.y=f[1];kk.step=t.step+1; w[0][f[1]]=1; Q.push (KK);} if(!w[g[0]][f[1]]) {man kk;kk.x=g[0];kk.y=f[1];kk.step=t.step+1; w[g[0]][f[1]]=1; Q.push (KK);}  for(intI=0; i<2; i++) {f[0]=T.x; f[1]=T.y; if(f[i]==0)Continue;  for(intj=0; j<2; J + +) {f[0]=T.x; f[1]=T.y;                Man K; if(i==j| | F[J]==G[J])Continue; if(f[i]+f[j]>=G[j]) {F[i]=f[i]-(g[j]-F[j]); F[J]=G[j]; //printf ("@%d%d\n", F[i],f[j]);}Else if(f[i]+f[j]<G[j]) {F[j]+=F[i]; F[i]=0; } k.x=f[0]; K.y=f[1]; //printf ("!!! %d%d%d\n ", k.x,k.y,k.z);                if(w[k.x][k.y]==0) {K.step=t.step+1;                    Q.push (k); W[K.X][K.Y]=1; }            }        }    }}intMain () {memset (W,0,sizeof(w)); CIN>>X>>Y>>K>>M; g[0]=y; g[1]=x;   BFS (); cout<<minn<<Endl; return 0;}

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Milk pails (BFS)