HDU Search Practice very Coke

Very Coke Time limit:2000/1000ms (java/other) Memory limit:32768/32768k (Java/other) total submission (s): Wuyi Accepted Submis Sion (s): 21Problem description It is a pleasant thing to drink coke after exercise, but Seeyou doesn't think so. Because every time when Seeyou bought Coke, Ox asked to share this bottle of Coke with seeyou, and must drink as much as seeyou. But Seeyou's hands only two cups, their capacity is N ml and M ml cola volume for S (s<101) ml (just fill a bottle), they can pour coke between three each other (there is no scale, and s==n+m,101>s>0,n>0,m>0). Smart Acmer, do you think they can split it? If you can output the minimum number of Coke, if not output "NO".

Input three integers: the volume of S Cola, N and M is the capacity of two cups, ending with "0 0 0".

Output if it can be divided into a minimum number of times to pour, otherwise output "NO".

Sample Input7 4 34 1 30 0 0

Sample OutputNO3 Simple test instructions: the title meaning is pour coke, divided into the same points, first judge can be the same, and then ask the least to how many times thinking analysis: Breadth search, have the goal, until the Pour Cola, to achieve the goal so far

#include <iostream>#include<algorithm>#include<queue>#include<cstring>using namespaceStd//according to the coursewareBOOL  is[101][101];intm, n, S, si, SJ;structnode{intX,y,all,t;//X,y,all represents the volume of Coke in the M,n,s Cup, and T indicates how many times it has been poured};voidBFs () {Queue<node>que; memset ( is,false,sizeof( is));    Node p,q; P.x=0, p.y =0, p.t =0, P.all =s;    Que.push (P);  is[P.x] [P.Y] =true;  while(!Que.empty ()) {P=Que.front ();        Que.pop (); if(p.y = = P.all && P.y = = s/2) {cout<< p.t <<Endl; return; }        if(P.all+p.x > m)//s Inverted m{q.x= M,q.y = P.y,q.all = p.all+p.x-m,q.t = p.t+1; if(! is[q.x][q.y]) Que.push (q), is[Q.x] [Q.Y] =true; }        Else{q.x= P.all+p.x,q.y = P.y,q.all =0, q.t = p.t+1; if(! is[q.x][q.y]) Que.push (q), is[Q.x] [Q.Y] =true; }        if(P.all+p.y > N)//s pour n{q.x= p.x, Q.y = n, q.all = p.all+p.y-n,q.t = p.t+1; if(! is[q.x][q.y]) Que.push (q), is[Q.x] [Q.Y] =true; }        Else{q.x= P.x,q.y = P.all+p.y,q.all =0, q.t = p.t+1; if(! is[q.x][q.y]) Que.push (q), is[Q.x] [Q.Y] =true; }        if(P.x+p.y > N)//m pour n{q.x= P.x+p.y-n,q.y = N,q.all = p.all,q.t = p.t+1; if(! is[q.x][q.y]) Que.push (q), is[Q.x] [Q.Y] =true; }        Else{q.x=0, q.y = P.x+p.y,q.all = p.all,q.t = p.t+1; if(! is[q.x][q.y]) Que.push (q), is[Q.x] [Q.Y] =true; } Q.all= p.all+p.x,q.x =0, q.y = p.y,q.t = p.t+1;//m Inverted S        if(! is[q.x][q.y]) Que.push (q), is[Q.x] [Q.Y] =true; if(P.x+p.y >m) {q.y= p.y+p.x-m,q.x = M,q.all = p.all,q.t = p.t+1;//N Inverted m            if(! is[q.x][q.y]) Que.push (q), is[Q.x] [Q.Y] =true; }        Else{q.x= P.X+P.Y,Q.Y =0, Q.all = p.all,q.t = p.t+1; if(! is[q.x][q.y]) Que.push (q), is[Q.x] [Q.Y] =true; } Q.all= p.all+p.y,q.x = P.x,q.y =0, q.t = p.t+1;//N Inverted S        if(! is[q.x][q.y]) Que.push (q), is[Q.x] [Q.Y] =true; } cout<<"NO"<<Endl;}intMain () { while(Cin >> s >> m >>N) {if(s = =0&& m = =0&& n = =0)         Break; if(s%2) {cout<<"NO"<<Endl; Continue; }        if(M >N) swap (m,n);    BFS (); }    return 0;}

　　

HDU Search Practice very Coke