POJ-3414.Pots.(BFS + 路径打印)

　　这道题做了很长时间，一开始上课的时候手写代码，所以想到了很多细节，但是创客手打代码的时候由于疏忽又未将pair赋初值，导致一直输出错误，以后自己写代码可以专心一点，可能会在宿舍图书馆或者Myhome，创客晚上好吵呀，隔壁真的服...

　　本题大意：给定两个杯子的容量，有六种操作，通过操作使得两个被子其中一个杯子的水等于待输入值C，并输出最少操作次数和操作名称。

　　本题思路：BFS，遇到合适的直接输出，搜完都没有搜到就impossible。

　　参考代码：

 　　将switch改为if能节省大部分篇幅，但是switch更美观明了一点。

 1 #include <string>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <queue>
 5 #include <stack>
 6 using namespace std;
 7 
 8 typedef pair<int ,int > P;
 9 const int maxn = 100 + 5, INF = 0x3f3f3f3f;
10 int A, B, C, t;
11 
12 struct {
13     int x, y, cnt;//用来存他爹的坐标和倒水的方式
14 } Ans[maxn][maxn];
15 int vis[maxn][maxn];
16 string ans[6] = {
17     "FILL(1)", "FILL(2)", "DROP(1)", "DROP(2)", "POUR(1,2)", "POUR(2,1)",
18 };
19 
20 void bfs() {
21     queue <P> Q;
22     vis[0][0] = 0;
23     Ans[0][0].x = -1, Ans[0][0].y = -1;
24     Q.push(make_pair(0, 0));
25     while(!Q.empty()) {
26         P now = Q.front(), Next;
27         stack <string> s1;
28         Q.pop();
29         if(now.first == C || now.second == C) {
30             cout << vis[now.first][now.second] << endl;
31             int X = now.first;
32             int Y = now.second;
33             while(X != -1) {
34                 s1.push(ans[Ans[X][Y].cnt]);
35                 int tmp = X;
36                 X = Ans[X][Y].x;
37                 Y = Ans[tmp][Y].y;
38             }
39             s1.pop();
40             while(!s1.empty()) {
41                 cout << s1.top() << endl;
42                 s1.pop();
43             }
44             return;
45         }
46         for(int i = 0; i < 6; i ++) {
47             switch(i) {
48                 case 0 ://将A加满
49                    Next.first = A;
50                    Next.second = now.second;
51                    break;
52                 case 1 ://将B加满
53                     Next.second = B;
54                     Next.first = now.first;
55                     break;
56                 case 2 ://将A倒掉
57                     Next.second = now.second;
58                     Next.first = 0;
59                     break;
60                 case 3 ://将B倒掉
61                     Next.first = now.first;
62                     Next.second = 0;
63                     break;
64                 case 4 ://将A倒入B中
65                     t = B - now.second;//B中还能倒入的量
66                     Next.first = now.first - t;
67                     if(Next.first < 0) Next.first = 0;
68                     Next.second = now.first + now.second;
69                     if(Next.second > B) Next.second = B;
70                     break;
71                 case 5 ://将B倒入A中
72                     t = A - now.first;//A中还能倒入的量
73                     Next.second = now.second  - t;
74                     if(Next.second < 0) Next.second = 0;
75                     Next.first = now.first + now.second;
76                     if(Next.first > A) Next.first = A;
77                     break;
78             }
79             if(vis[Next.first][Next.second] == INF) {
80                 vis[Next.first][Next.second] = vis[now.first][now.second] + 1;
81                 Q.push(Next);
82                 Ans[Next.first][Next.second].x = now.first;
83                 Ans[Next.first][Next.second].y = now.second;
84                 Ans[Next.first][Next.second].cnt = i;
85             }
86         }
87     }
88     cout << "impossible" << endl;
89 }
90 
91 int main () {
92     memset(vis, INF, sizeof(vis));
93     cin >> A >> B >> C;
94     bfs();
95     return 0;
96 }

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