poj1606 Jugs（BFS）

题目链接

题意

有两个容量分别为ca,cb的杯子，可以向杯子里倒水，将杯子里的水倒空，将一个杯子里的水倒到另一个杯子里，求怎样倒才能使其中的一个杯子里的水恰为N加仑。

思路

两个杯子里的水量组成一个状态，不断地进行题中的6种操作来扩展状态结点进行bfs，直到其中一个杯子的水量为N即可。

代码

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <queue>
  5 #include <stack>
  6 using namespace std;
  7 
  8 struct Node
  9 {
 10     int a, b;
 11     int flag;
 12     Node* pre;
 13 };
 14 
 15 const int N = 1010;
 16 int ca, cb, n;
 17 int visit[N][N];
 18 stack<int> s;
 19 
 20 void print()
 21 {
 22     while (!s.empty())
 23     {
 24         switch (s.top())
 25         {
 26         case 0:
 27             cout << "fill A" << endl;
 28             break;
 29         case 1:
 30             cout << "fill B" << endl;
 31             break;
 32         case 2:
 33             cout << "empty A" << endl;
 34             break;
 35         case 3:
 36             cout << "empty B" << endl;
 37             break;
 38         case 4:
 39             cout << "pour A B" << endl;
 40             break;
 41         case 5:
 42             cout << "pour B A" << endl;
 43             break;
 44         }
 45         s.pop();
 46     }
 47     cout << "success" << endl;
 48 }
 49 
 50 void bfs(int a, int b)
 51 {
 52     Node state[N];
 53     int cnt = -1;
 54     memset(visit, 0, sizeof(visit));
 55     Node node;
 56     node.a = node.b = 0;
 57     node.pre = NULL;
 58     queue<Node> q;
 59     q.push(node);
 60     visit[node.a][node.b] = 1;
 61     while (!q.empty())
 62     {
 63         Node node = q.front();
 64         q.pop();
 65         state[++cnt] = node;
 66         Node next = node;
 67         for (int i = 0; i < 6; i++)
 68         {
 69             next = node;
 70             int amount;
 71             switch (i)
 72             {
 73             case 0:        //fill a
 74                 next.a = ca;
 75                 next.flag = 0;
 76                 break;
 77             case 1:        //fill b
 78                 next.b = cb;
 79                 next.flag = 1;
 80                 break;
 81             case 2:        // empty a
 82                 next.a = 0;
 83                 next.flag = 2;
 84                 break;
 85             case 3:        //empty b
 86                 next.b = 0;
 87                 next.flag = 3;
 88                 break;
 89             case 4:        //pour a b
 90                 amount = cb - node.b;
 91                 if (node.a > amount)
 92                 {
 93                     next.a -= amount;
 94                     next.b = cb;
 95                 }
 96                 else {
 97                     next.a = 0;
 98                     next.b = node.a + node.b;
 99                 }
100                 next.flag = 4;
101                 break;
102             case 5:        //pour b a
103                 amount = ca - node.a;
104                 if (node.b > amount)
105                 {
106                     next.a = ca;
107                     next.b -= amount;
108                 }
109                 else {
110                     next.a = node.a + node.b;
111                     next.b = 0;
112                 }
113                 next.flag = 5;
114                 break;
115             }
116 
117             if (!visit[next.a][next.b])
118             {
119                 visit[next.a][next.b] = 1;
120                 next.pre = &state[cnt];
121                 if (next.a==n || next.b == n)
122                 {
123                     while (next.pre)
124                     {
125                         s.push(next.flag);
126                         next = *next.pre;
127                     }
128                     print();
129                     return;
130                 }
131                 q.push(next);
132             }
133         }
134     }
135 }
136 
137 int main()
138 {
139     //freopen("poj1606.txt", "r", stdin);
140     while (cin >> ca >> cb >> n)
141     {
142         bfs(0, 0);
143     }
144     return 0;
145 }

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