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 3 4 1 3 0 0 0
Sample Output
NO 3
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 99999999
using 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;
}