HDU 1495 非常可乐

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

思路

六种倒水情况，a -> b, a -> c, b -> a, b -> c, c -> a, c -> b, BFS穷竭搜索

Code

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

int s, n, m;
struct node
{
    int a, b, c, step;
};
bool vis[105][105][105];

int bfs()
{
    node now;
    now.a = s;
    now.b = 0;
    now.c = 0;
    now.step = 0;
    queue<node> q;
    q.push(now);
    vis[s][0][0] = 1;
    while(q.size())
    {
        now = q.front();
        q.pop();
        if((now.a == 0 && now.b == now.c) || (now.b == 0 && now.a == now.c) || (now.c == 0 && now.a == now.b))
            return now.step;
        node nex;
        int tra;///要转移的水
        if(now.a)///a剩有部分水
        {
            tra = min(now.a, n - now.b);///a -> b 转移的水要么是a所剩，要么是b所缺
            if(tra)///这部分 > 0 才可转移
            {
                nex.a = now.a - tra;
                nex.b = now.b + tra;
                nex.c = now.c;
                nex.step = now.step + 1;
                if(!vis[nex.a][nex.b][nex.c])
                {
                    q.push(nex);
                    vis[nex.a][nex.b][nex.c] = 1;
                }
            }
            tra = min(now.a, m - now.c);///a -> c
            if(tra)
            {
                nex.a = now.a - tra;
                nex.c = now.c + tra;
                nex.b = now.b;
                nex.step = now.step + 1;
                if(!vis[nex.a][nex.b][nex.c])
                {
                    q.push(nex);
                    vis[nex.a][nex.b][nex.c] = 1;
                }
            }
        }
        if(now.b)
        {
            tra = min(now.b, s - now.a);///b -> a
            if(tra)
            {
                nex.b = now.b - tra;
                nex.a = now.a + tra;
                nex.c = now.c;
                nex.step = now.step + 1;
                if(!vis[nex.a][nex.b][nex.c])
                {
                    q.push(nex);
                    vis[nex.a][nex.b][nex.c] = 1;
                }
            }
            tra = min(now.b, m - now.c);///b -> c
            if(tra)
            {
                nex.b = now.b - tra;
                nex.c = now.c + tra;
                nex.a = now.a;
                nex.step = now.step + 1;
                if(!vis[nex.a][nex.b][nex.c])
                {
                    q.push(nex);
                    vis[nex.a][nex.b][nex.c] = 1;
                }
            }
        }
        if(now.c)
        {
            tra = min(now.c, s - now.a);///c -> a
            if(tra)
            {
                nex.c = now.c - tra;
                nex.a = now.a + tra;
                nex.b = now.b;
                nex.step = now.step + 1;
                if(!vis[nex.a][nex.b][nex.c])
                {
                    q.push(nex);
                    vis[nex.a][nex.b][nex.c] = 1;
                }
            }
            tra = min(now.c, n - now.b);///c -> b
            if(tra)
            {
                nex.c = now.c - tra;
                nex.b = now.b + tra;
                nex.a = now.a;
                nex.step = now.step + 1;
                if(!vis[nex.a][nex.b][nex.c])
                {
                    q.push(nex);
                    vis[nex.a][nex.b][nex.c] = 1;
                }
            }
        }
    }
    return 0;
}

int main()
{
    while(~scanf("%d%d%d", &s, &n, &m) &&(s || n || m))
    {
        if(s & 1)///奇数不可能平分
            cout << "NO" << '\n';
        else
        {
            memset(vis, 0, sizeof(vis));
            int ans = bfs();
            if(ans)
                cout << ans << '\n';
            else
                cout << "NO" << '\n';
        }
    }
    return 0;
}