題目鏈接

http://poj.org/problem?id=3414

題意

有兩個杯子，容量分別為A升，B升，可以向杯子裏倒滿水，將杯子裏的水倒空，將一個杯子裏的水倒到另一個杯子裏，求怎樣倒才能使其中的一個杯子裏的水恰為C升，輸出最少步數和操作；如果不能倒到C升，輸出“impossible”。

思路

這題與poj1606基本相同，在poj1606的基礎上添加了輸出最少步數，修改了操作的表示，以及不可能達到目標時輸出impossible。將poj1606的代碼略作修改即可。

代碼

  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 steps;
 12     int flag;
 13     Node* pre;
 14 };
 15 
 16 const int N = 1010;
 17 int ca, cb, n;
 18 int visit[N][N];
 19 stack<int
 
> s;
 20 
 21 void print()
 22 {
 23     while (!s.empty())
 24     {
 25         switch (s.top())
 26         {
 27         case 0:
 28             cout << "FILL(1)" << endl;
 29             break;
 30         case 1:
 31             cout << "FILL(2)" << endl;
 32
 
             break;
 33         case 2:
 34             cout << "DROP(1)" << endl;
 35             break;
 36         case 3:
 37             cout << "DROP(2)" << endl;
 38             break;
 39         case 4:
 40             cout << "POUR(1,2)" << endl;
 41             break;
 42         case 5:
 43             cout << "POUR(2,1)" << endl;
 44             break;
 45         }
 46         s.pop();
 47     }
 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.steps = 0;
 58     node.pre = NULL;
 59     queue<Node> q;
 60     q.push(node);
 61     visit[node.a][node.b] = 1;
 62     while (!q.empty())
 63     {
 64         Node node = q.front();
 65         q.pop();
 66         state[++cnt] = node;
 67         Node next = node;
 68         for (int i = 0; i < 6; i++)
 69         {
 70             next = node;
 71             int amount;
 72             switch (i)
 73             {
 74             case 0:        //FILL(1)
 75                 next.a = ca;
 76                 next.flag = 0;
 77                 break;
 78             case 1:        //FILL(2)
 79                 next.b = cb;
 80                 next.flag = 1;
 81                 break;
 82             case 2:        // DROP(1)
 83                 next.a = 0;
 84                 next.flag = 2;
 85                 break;
 86             case 3:        //DROP(2)
 87                 next.b = 0;
 88                 next.flag = 3;
 89                 break;
 90             case 4:        //POUR(1,2)
 91                 amount = cb - node.b;
 92                 if (node.a > amount)
 93                 {
 94                     next.a -= amount;
 95                     next.b = cb;
 96                 }
 97                 else {
 98                     next.a = 0;
 99                     next.b = node.a + node.b;
100                 }
101                 next.flag = 4;
102                 break;
103             case 5:        //POUR(2,1)
104                 amount = ca - node.a;
105                 if (node.b > amount)
106                 {
107                     next.a = ca;
108                     next.b -= amount;
109                 }
110                 else {
111                     next.a = node.a + node.b;
112                     next.b = 0;
113                 }
114                 next.flag = 5;
115                 break;
116             }
117 
118             if (!visit[next.a][next.b])
119             {
120                 visit[next.a][next.b] = 1;
121                 next.pre = &state[cnt];
122                 next.steps = node.steps + 1;
123                 if (next.a == n || next.b == n)
124                 {
125                     cout << next.steps << endl;
126                     while (next.pre)
127                     {
128                         s.push(next.flag);
129                         next = *next.pre;
130                     }
131                     print();
132                     return;
133                 }
134                 q.push(next);
135             }
136         }
137     }
138     cout << "impossible" << endl;
139 }
140 
141 int main()
142 {
143     cin >> ca >> cb >> n;
144     bfs(0, 0);
145     return 0;
146 }

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