POJ_3414 Pots 【复杂BFS】

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <queue>
  5 using namespace std;
  6 const int MAXN = 100;
  7 bool visit[MAXN+3][MAXN+3];
  8 int A, B, C;
  9 
 10 struct Point
 11 {
 12     int first, second, cnt;
 13     int prev, id;   //父节点和操作对象
 14     char op; //操作
 15 };
 16 
 17 Point P[MAXN*MAXN + 4];
 18 int Cnt, Ans;
 19 
 20 void Output(int t)
 21 {
 22     if(P[t].prev != -1)
 23     {
 24         Ans++;
 25         Output(P[t].prev);
 26     }
 27     if(Ans != -1)
 28     {
 29         printf("%d\n", Ans);
 30         Ans = -1;
 31     }
 32     if(P[t].op==‘F‘)
 33     {
 34         printf("FILL(%d)\n", P[t].id);
 35     }
 36     else if(P[t].op == ‘P‘)
 37     {
 38         printf("POUR(%d,%d)\n", P[t].id, P[t].id==1?2:1);
 39     }
 40     else if(P[t].op == ‘D‘)
 41     {
 42         printf("DROP(%d)\n", P[t].id);
 43     }
 44 }
 45 
 46 void BFS()
 47 {
 48     Point t;
 49     int cur;
 50     t.first = 0, t.second = 0;
 51     visit[0][0] = 1;
 52     t.op = ‘0‘, t.prev = -1, t.id = -1;
 53     t.cnt = 0;
 54     P[0] = t;
 55     Cnt = 1;
 56     cur = 0;
 57 
 58     while(true)
 59     {
 60         if(cur >= Cnt)
 61         {
 62             printf("impossible\n");
 63             return;
 64         }
 65         Point pt = P[cur++];
 66 
 67         if(pt.first == C || pt.second == C)
 68         {
 69             Ans = 0;
 70             Output(pt.cnt);
 71             break;
 72         }
 73 
 74         t.prev = pt.cnt;
 75 
 76         if(pt.first < A)
 77         {
 78             t.first = A;
 79             t.second = pt.second;
 80             if(visit[t.first][t.second] == 0)
 81             {
 82                 //visit[t.first][t.second] == 1; 刚开始RE的原因
 83                 visit[t.first][t.second] = 1;
 84                 t.op = ‘F‘;
 85                 t.id = 1;
 86                 t.cnt = Cnt;
 87                 P[Cnt++] = t;
 88 
 89             }
 90         }
 91 
 92         if(pt.second < B)
 93         {
 94             t.first = pt.first;
 95             t.second = B;
 96             if(visit[t.first][t.second] == 0)
 97             {
 98                 visit[t.first][t.second] = 1;
 99                 t.op = ‘F‘;
100                 t.id = 2;
101                 t.cnt = Cnt;
102                 P[Cnt++] = t;
103             }
104         }
105 
106         if(pt.first < A && pt.second > 0 )
107         {
108             t.first = pt.first + pt.second;
109             t.second = t.first - A;
110             if(t.second < 0)
111                 t.second = 0;
112             else
113                 t.first = A;
114             if(visit[t.first][t.second] == 0)
115             {
116                 visit[t.first][t.second] = 1;
117                 t.op = ‘P‘;
118                 t.id = 2;
119                 t.cnt = Cnt;
120                 P[Cnt++] = t;
121             }
122         }
123 
124         if(pt.second < B && pt.first > 0)
125         {
126             t.second = pt.second + pt.first;
127             t.first = t.second - B;
128             if(t.first < 0)
129                 t.first = 0;
130             else
131                 t.second = B;
132             if(visit[t.first][t.second] == 0)
133             {
134                 visit[t.first][t.second] = 1;
135                 t.op = ‘P‘;
136                 t.id = 1;
137                 t.cnt = Cnt;
138                 P[Cnt++] = t;
139             }
140         }
141 
142         if(pt.first > 0)
143         {
144             t.first = 0;
145             t.second = pt.second;
146             if(visit[t.first][t.second] == 0)
147             {
148                 visit[t.first][t.second] = 1;
149                 t.op = ‘D‘;
150                 t.id = 1;
151                 t.cnt = Cnt;
152                 P[Cnt++] = t;
153             }
154         }
155 
156         if(pt.second > 0)
157         {
158             t.first = pt.first;
159             t.second = 0;
160             if(visit[t.first][t.second] == 0)
161             {
162                 visit[t.first][t.second] = 1;
163                 t.op = ‘D‘;
164                 t.id = 2;
165                 t.cnt = Cnt;
166                 P[Cnt++] = t;
167             }
168         }
169     }
170 }
171 
172 int main()
173 {
174     while(scanf("%d %d %d", &A, &B, &C)!=EOF)
175     {
176         memset(visit, 0, sizeof(visit));
177         BFS();
178     }
179     return 0;
180 }