鏈接：https://vjudge.net/problem/HDU-1495

題意：

大家一定覺的運動以後喝可樂是一件很愜意的事情，但是seeyou卻不這麽認為。因為每次當seeyou買了可樂以後，阿牛就要求和seeyou一起分享這一瓶可樂，而且一定要喝的和seeyou一樣多。但seeyou的手中只有兩個杯子，它們的容量分別是N 毫升和M 毫升 可樂的體積為S （S<101）毫升　(正好裝滿一瓶) ，它們三個之間可以相互倒可樂 (都是沒有刻度的，且 S==N+M，101＞S＞0，N＞0，M＞0) 。聰明的ACMER你們說他們能平分嗎？如果能請輸出倒可樂的最少的次數，如果不能輸出"NO"。

思路：

bfs，每次有六種操作，挨個嘗試，當兩個杯子的值是總水量的一半時返回步數。

否則返回-1。

代碼：

#include <iostream>
#include <memory.h>
#include <vector>
#include <map>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <queue>
#include <string>

using namespace std;

typedef long long LL;

const int MAXN = 100 + 10;

struct Node
{
    int s_all, s_used;
    int n_all, n_used;
    int m_all, m_used;
    int step;
    Node(int s1, int s2, int n1, int n2, int m1, int m2, int steps)
    {
        s_all = s1;
        s_used = s2;
        n_all = n1;
        n_used = n2;
        m_all = m1;
        m_used = m2;
        step = steps;
    }
};
int s, n, m;
int vis[MAXN][MAXN][MAXN];

int Bfs(int s1, int n1, int m1)
{
    queue<Node> que;
    que.emplace(s, s, n, 0, m, 0, 0);
    vis[s1][0][0] = 1;
    while (!que.empty())
    {
        Node now = que.front();
        if (now.s_used == s / 2 && now.n_used == s / 2)
            return now.step;
        if (now.s_used == s / 2 && now.m_used == s / 2)
            return now.step;
        if (now.n_used == s / 2 && now.m_used == s / 2)
            return now.step;
        que.pop();
        int ns, nn, nm;
        ns = now.s_used - min(now.s_used, now.n_all - now.n_used);
        nn = now.n_used + min(now.s_used, now.n_all - now.n_used);
        nm = now.m_used;
        if (vis[ns][nn][nm] == 0)
        {
            que.emplace(now.s_all, ns, now.n_all, nn, now.m_all, nm, now.step + 1);
            vis[ns][nn][nm] = 1;
        }
        ns = now.s_used - min(now.s_used, now.m_all - now.m_used);
        nn = now.n_used;
        nm = now.m_used + min(now.s_used, now.m_all - now.m_used);
        if (vis[ns][nn][nm] == 0)
        {
            que.emplace(now.s_all, ns, now.n_all, nn, now.m_all, nm, now.step + 1);
            vis[ns][nn][nm] = 1;
        }
        ns = now.s_used + min(now.n_used, now.s_all - now.s_used);
        nn = now.n_used - min(now.n_used, now.s_all - now.s_used);
        nm = now.m_used;
        if (vis[ns][nn][nm] == 0)
        {
            que.emplace(now.s_all, ns, now.n_all, nn, now.m_all, nm, now.step + 1);
            vis[ns][nn][nm] = 1;
        }
        ns = now.s_used;
        nn = now.n_used - min(now.n_used, now.m_all - now.m_used);
        nm = now.m_used + min(now.n_used, now.m_all - now.m_used);
        if (vis[ns][nn][nm] == 0)
        {
            que.emplace(now.s_all, ns, now.n_all, nn, now.m_all, nm, now.step + 1);
            vis[ns][nn][nm] = 1;
        }
        ns = now.s_used + min(now.m_used, now.s_all - now.s_used);
        nn = now.n_used;
        nm = now.m_used - min(now.m_used, now.s_all - now.s_used);
        if (vis[ns][nn][nm] == 0)
        {
            que.emplace(now.s_all, ns, now.n_all, nn, now.m_all, nm, now.step + 1);
            vis[ns][nn][nm] = 1;
        }
        ns = now.s_used;
        nn = now.n_used + min(now.m_used, now.n_all - now.n_used);
        nm = now.m_used - min(now.m_used, now.n_all - now.n_used);
        if (vis[ns][nn][nm] == 0)
        {
            que.emplace(now.s_all, ns, now.n_all, nn, now.m_all, nm, now.step + 1);
            vis[ns][nn][nm] = 1;
        }
    }
    return -1;
}

int main()
{
    while (cin >> s >> n >> m)
    {
        memset(vis, 0, sizeof(vis));
        if (s == 0 && n == 0 && m == 0)
            break;
        if (s % 2 != 0)
        {
            cout << "NO" << endl;
            continue;
        }
        int res = Bfs(s, n, m);
        if (res != -1)
            cout << res << endl;
        else
            cout << "NO" << endl;
    }

    return 0;
}

HDU-1495-非常可樂