hdu 1495(BFS)

非常可乐

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11034    Accepted Submission(s): 4458

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

 

Author

seeyou

 

题解:奇数的除外,总共六种倒法,a->b a->c b->a b->c c->a c->b.三种状态,仔细一点模拟就可以做出来了。

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <stdlib.h>
using namespace std;

struct Node{
    int x,y,z,step;
};
int a,b,c;
bool vis[105][105][105];
int bfs(){
    memset(vis,false,sizeof(vis));
    queue<Node> q;
    Node s;
    s.x = a,s.y = 0,s.z = 0,s.step = 0;
    vis[a][0][0]=true;
    q.push(s);
    while(!q.empty()){
        Node now = q.front();
        q.pop();
        if(now.x==a/2&&now.y==a/2||now.x==a/2&&now.z==a/2||now.y==a/2&&now.z==a/2){
            return now.step;
        }
        Node next;
        if(now.x!=0){  /// a -> b
            if(now.x>b-now.y){ ///还可以倒满b
                next.x =now.x-(b-now.y);
                next.y = b;
                next.z = now.z;
            }else{
                next.x = 0;
                next.y =now.x+now.y;
                next.z = now.z;
            }
            next.step=now.step+1;
            if(!vis[next.x][next.y][next.z]){
                vis[next.x][next.y][next.z] = true;
                q.push(next);
            }
        }
        if(now.x!=0){ ///a->c
            if(now.x>c-now.z){
                next.x = now.x - (c-now.z);
                next.y = now.y;
                next.z = c;
            }else{
                next.x = 0;
                next.y = now.y;
                next.z =now.z+now.x;
            }
            next.step=now.step+1;
            if(!vis[next.x][next.y][next.z]){
                vis[next.x][next.y][next.z] = true;
                q.push(next);
            }
        }
        if(now.y!=0){ ///b->a
            if(now.y>a-now.x){
                next.x = a;
                next.y = now.y - (a-now.x);
                next.z = now.z;
            }else{
                next.x = now.x+now.y;
                next.y = 0;
                next.z =now.z;
            }
           next.step=now.step+1;
            if(!vis[next.x][next.y][next.z]){
                vis[next.x][next.y][next.z] = true;
                q.push(next);
            }
        }
        if(now.y!=0){ ///b->c
            if(now.y>c-now.z){
                next.x = now.x;
                next.y = now.y - (c-now.z);
                next.z = c;
            }else{
                next.x = now.x;
                next.y = 0;
                next.z =now.z+now.y;
            }
            next.step=now.step+1;
            if(!vis[next.x][next.y][next.z]){
                vis[next.x][next.y][next.z] = true;
                q.push(next);
            }
        }
        if(now.z!=0){ ///c->a
            if(now.z>a-now.x){
                next.x = a;
                next.y = now.y;
                next.z = now.z - (a-now.x);
            }else{
                next.x = now.x+now.z;
                next.y = now.y;
                next.z = 0;
            }
            next.step=now.step+1;
            if(!vis[next.x][next.y][next.z]){
                vis[next.x][next.y][next.z] = true;
                q.push(next);
            }
        }
        if(now.z!=0){ ///c->b
            if(now.z>b-now.y){
                next.x = now.x;
                next.y = b;
                next.z = now.z - (b-now.y);
            }else{
                next.x = now.x;
                next.y = now.y+now.z;
                next.z = 0;
            }
            next.step=now.step+1;
            if(!vis[next.x][next.y][next.z]){
                vis[next.x][next.y][next.z] = true;
                q.push(next);
            }
        }
    }
    return -1;
}
int main(){
    while(scanf("%d%d%d",&a,&b,&c)!=EOF,a+b+c){
        if(a%2==1){
            printf("NO
");
            continue;
        }
        int ans = bfs();
        if(ans==-1){
            printf("NO
");
        }else{
            printf("%d
",ans);
        }
    }
}