hdu1495之经典搜索

非常可乐Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3688    Accepted Submission(s): 1533


Problem Description大家一定觉的运动以后喝可乐是一件很惬意的事情，但是seeyou却不这么认为。因为每次当seeyou买了可乐以后，阿牛就要求和seeyou一起分享这一瓶可乐，而且一定要喝的和seeyou一样多。但seeyou的手中只有两个杯子，它们的容量分别是N 毫升和M 毫升 可乐的体积为S （S<101）毫升　(正好装满一瓶) ，它们三个之间可以相互倒可乐 (都是没有刻度的，且 S==N+M，101＞S＞0，N＞0，M＞0) 。聪明的ACMER你们说他们能平分吗？如果能请输出倒可乐的最少的次数，如果不能输出"NO"。 
Input三个整数 : S 可乐的体积 , N 和 M是两个杯子的容量，以"0 0 0"结束。 
Output如果能平分的话请输出最少要倒的次数，否则输出"NO"。 
Sample Input

7 4 34 1 30 0 0

 
Sample Output

NO3

分析：因为有三个杯子a,b,c,所以对于下一步可以分6种情况:a向b,c中倒水,b向a,c中倒水,c向a,b中倒水,对这6种情况直接搜索即可

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<queue>#include<algorithm>#include<map>#include<iomanip>#define INF 99999999using namespace std;const int MAX=101+10;int sum,n,m;bool mark[MAX][MAX];//表示杯子a,b有水i,j的状态是否到达过 struct Node{int a,b,c,time;//代表杯子a,b,c里的水量和倒水的次数 Node(){}Node(int A,int B,int C,int Time):a(A),b(B),c(C),time(Time){} }start;int BFS(){memset(mark,false,sizeof mark);start=Node(2*sum,0,0,0);mark[start.a][start.b]=true;Node oq,next;queue<Node>q;q.push(start);while(!q.empty()){oq=q.front();q.pop();int x=2*sum-oq.a,y=n-oq.b,z=m-oq.c;if(x){//向a里面可以倒水if(oq.b)next=Node(oq.a+min(x,oq.b),oq.b-min(x,oq.b),oq.c,oq.time+1);//b向a倒水 if(!mark[next.a][next.b]){mark[next.a][next.b]=true;if(next.a == sum && next.b == sum || (next.a == sum && next.c == sum) || (next.b == sum && next.c == sum))return next.time;q.push(next);}if(oq.c)next=Node(oq.a+min(x,oq.c),oq.b,oq.c-min(x,oq.c),oq.time+1);//c向a倒水 if(!mark[next.a][next.b]){mark[next.a][next.b]=true;if(next.a == sum && next.b == sum || (next.a == sum && next.c == sum) || (next.b == sum && next.c == sum))return next.time;q.push(next);}}if(y){//向b里面倒水 if(oq.a)next=Node(oq.a-min(oq.a,y),oq.b+min(oq.a,y),oq.c,oq.time+1);//a向b倒水if(!mark[next.a][next.b]){mark[next.a][next.b]=true;if(next.a == sum && next.b == sum || (next.a == sum && next.c == sum) || (next.b == sum && next.c == sum))return next.time;q.push(next);}if(oq.c)next=Node(oq.a,oq.b+min(y,oq.c),oq.c-min(y,oq.c),oq.time+1);//c向b倒水if(!mark[next.a][next.b]){mark[next.a][next.b]=true;if(next.a == sum && next.b == sum || (next.a == sum && next.c == sum) || (next.b == sum && next.c == sum))return next.time;q.push(next);}}if(z){//向c倒水 if(oq.a)next=Node(oq.a-min(oq.a,z),oq.b,oq.c+min(oq.a,z),oq.time+1);//a向c倒水if(!mark[next.a][next.b]){mark[next.a][next.b]=true;if(next.a == sum && next.b == sum || (next.a == sum && next.c == sum) || (next.b == sum && next.c == sum))return next.time;q.push(next);}if(oq.b)next=Node(oq.a,oq.b-min(oq.b,z),oq.c+min(oq.b,z),oq.time+1);//b向c倒水if(!mark[next.a][next.b]){mark[next.a][next.b]=true;if(next.a == sum && next.b == sum || (next.a == sum && next.c == sum) || (next.b == sum && next.c == sum))return next.time;q.push(next);}}}return -1;}int main(){while(cin>>sum>>n>>m,sum+n+m){if(sum%2){cout<<"NO"<<endl;continue;}sum=sum/2;int temp=BFS();if(temp == -1)cout<<"NO"<<endl;else cout<<temp<<endl;}return 0;}