HDU-1495-非常可乐
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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 <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; }
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