HDU-1495-非常可乐-白红宇

HDU-1495-非常可乐

阅读量：6695 次

发布时间：2019-06-25

本文共 3119 字，大约阅读时间需要 10 分钟。

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

题意：

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"。

思路：

bfs，每次有六种操作，挨个尝试，当两个杯子的值是总水量的一半时返回步数。

否则返回-1。

代码：

#include 
     
      #include 
      
       #include 
       
        #include 
        #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              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
              
                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;}

转载于:https://www.cnblogs.com/YDDDD/p/10591362.html

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