poj 3414 pots

You are given two pots, having the volume of A and B liters respectively. The following operations can be performed:

  1. FILL(i)        fill the pot i (1 ≤ i ≤ 2) from the tap;
  2. DROP(i)      empty the pot i to the drain;
  3. POUR(i,j)    pour from pot i to pot j; after this operation either the pot j is full (and there may be some water left in the pot i), or the pot i is empty (and all its contents have been moved to the pot j).

Write a program to find the shortest possible sequence of these operations that will yield exactly C liters of water in one of the pots.

 

遇到的问题：

　　　　如何合理的拓展；分成8中情况一种种考虑;

 1 #include<cstring>
 2 #include<iostream>
 3 #include<cstdio>  
 4 #include<stack>  
 5 using namespace std;
 6 const int N = 111;
 7 bool vis[N][N];
 8 int A, B, C;
 9 struct state
10 {
11     int x, y, step, pre, id;
12     state() { }
13     state(int xx, int yy, int s, int p, int i): x(xx), y(yy), step(s), pre(p), id(i) { }
14 }q[N*N];
15 void print(state over) {
16     cout << over.step << endl;
17     stack<int> ans;
18     while (over.step) { 
19         ans.push(over.id);
20         over = q[over.pre];
21     }
22     while (!ans.empty()) {
23         int t = ans.top();
24         ans.pop();
25         switch (t)
26         {
27         case 1:
28             printf("FILL(1)\n"); break;
29         case 2:
30             printf("FILL(2)\n"); break;
31         case 3:
32             printf("DROP(1)\n"); break;
33         case 4:
34             printf("DROP(2)\n"); break;
35         case 5:
36             printf("POUR(1,2)\n"); break;
37         case 6:
38             printf("POUR(2,1)\n"); break;
39         }
40     }
41 }
42 bool bfs() {
43     memset(vis, 0, sizeof(vis));
44     int head = 0, tail = 0;
45     q[tail++] = state(0, 0, 0, -1, -1);
46     while (head < tail) {
47         state t = q[head];
48         int a = t.x, b = t.y, step = t.step;
49         if (a == C || b == C) {
50             print(t);
51             return true;
52         }
53         if (!vis[A][b]) {    //FILL(1)
54             vis[A][b] = 1;
55             q[tail++] = state(A, b, step + 1, head, 1);
56         }
57         if (!vis[a][B]) { //FILL(2)
58             vis[a][B] = 1;
59             q[tail++] = state(a, B, step + 1, head, 2);
60         }
61         if (!vis[0][b]) { //drop(1)
62             vis[0][b] = 1;
63             q[tail++] = state(0, b, step + 1, head, 3);
64         }
65         if (!vis[a][0]) { //drop(2)
66             vis[a][0] = 1;
67             q[tail++] = state(a, 0, step + 1, head, 4);
68         }
69         if (b + a > B && !vis[a - (B - b)][B]) {    //pour(1, 2) 1杯有剩余
70             vis[a - B + b][B] = 1;
71             q[tail++] = state(a - B + b, B, step + 1, head, 5);
72         }
73         if (b + a <= B && !vis[0][a + b]) {    //pour(1, 2) ， 1杯没有剩余;
74             vis[0][a + b] = 1;
75             q[tail++] = state(0, a + b, step + 1, head, 5);
76         }
77         if (b + a > A && !vis[A][b - A + a]) {    //pour(2, 1) 
78             vis[A][b - A + a] = 1;
79             q[tail++] = state(A, b - A + a, step + 1, head, 6);
80         }
81         if (b + a <= A && !vis[a + b][0]) { //2杯没有剩余;
82             vis[a + b][0] = 1;
83             q[tail++] = state(a + b, 0, step + 1, head, 6);
84         }
85         head++;
86     }
87     return false;
88 }
89 
90 int main() {
91     cin >> A >> B >> C;
92     if (!bfs())
93         cout << "impossible!" << endl;
94     return 0;
95 }
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posted @ 2018-03-13 22:15  Charlie_li  阅读(102)  评论(0)    收藏  举报