Pots
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 6116 |
|
Accepted: 2582 |
|
Special Judge |
Description
You are given two pots, having the volume of A and B liters respectively. The following operations can be performed:
- FILL(i) fill the pot i (1 ≤ i ≤ 2) from the tap;
- DROP(i) empty the pot i to the drain;
- 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.
Input
On the first and only line are the numbers A, B, and C. These are all integers in the range from 1 to 100 and C≤max(A,B).
Output
The first line of the output must contain the length of the sequence of operations K. The following K lines must each describe one operation. If there are several sequences of minimal length, output any one of them. If the desired result can’t be achieved, the first and only line of the file must contain the word ‘impossible’.
Sample Input
3 5 4
Sample Output
6
FILL(2)
POUR(2,1)
DROP(1)
POUR(2,1)
FILL(2)
POUR(2,1)
挺簡(jiǎn)單的題目
沒看完題目，不知道還有impossible的情況，wa了一次，然后第二遍忘了輸出impossible之后要return，re一次
呃，蛋疼
1
#include<stdio.h>
2#include<string.h>
3#include<math.h>
4int a,b,c;
5struct node
6
{
7
int a,b,d,d1;
8
};
9struct node q[100050];
10int wh[10005];
11int num0[10005],num;
12short flag[105][105];
13int head,tail,k;
14int bfs()
15{
16 int nowa,nowb,aa,bb;
17 head=0;
18 tail=1;
19 q[tail].a=0;
20 q[tail].b=0;
21 flag[0][0]=1;
22 wh[tail]=0;
23 while (head<tail)
24
{
25 head++;
26 nowa=q[head].a;
27 nowb=q[head].b;
28 if ((q[head].a==c)||(q[head].b==c))
29 {
30 return head;
31
}
32 if (nowa!=a)
33 {
34 if (!flag[a][nowb])
35 {
36 tail++;
37 q[tail].a=a;
38 q[tail].b=nowb;
39 q[tail].d=1;
40 q[tail].d1=1;
41 wh[tail]=head;
42 flag[a][nowb]=1;
43 }
44 }
45 if (nowb!=b)
46 {
47 if (!flag[nowa][b])
48 {
49 tail++;
50 q[tail].a=nowa;
51 q[tail].b=b;
52 q[tail].d=1;
53 q[tail].d1=2;
54 wh[tail]=head;
55 flag[nowa][b]=1;
56 }
57 }
58 if (nowa!=0)
59 {
60 if (!flag[0][nowb])
61 {
62 tail++;
63 q[tail].a=0;
64 q[tail].b=nowb;
65 q[tail].d=2;
66 q[tail].d1=1;
67 wh[tail]=head;
68 flag[0][nowb]=1;
69 }
70 }
71 if (nowb!=0)
72 {
73 if (!flag[nowa][0])
74 {
75 tail++;
76 q[tail].a=nowa;
77 q[tail].b=0;
78 q[tail].d=2;
79 q[tail].d1=2;
80 wh[tail]=head;
81 flag[nowa][0]=1;
82 }
83 }
84 if (nowa!=0)
85 {
86 if (nowa>=b-nowb)
87 {
88 aa=nowa+nowb-b;
89 //printf("aa=%d\n",aa);
90 bb=b;
91 if (!flag[aa][bb])
92 {
93 tail++;
94 q[tail].a=aa;
95 q[tail].b=bb;
96 q[tail].d=3;
97 q[tail].d1=1;
98 wh[tail]=head;
99 flag[aa][bb]=1;
100 }
101 }
102 else if(nowa<b-nowb)
103 {
104 aa=0;
105 bb=nowb+nowa;
106 if (!flag[aa][bb])
107 {
108 tail++;
109 q[tail].a=aa;
110 q[tail].b=bb;
111 q[tail].d=3;
112 q[tail].d1=1;
113 wh[tail]=head;
114 flag[aa][bb]=1;
115 }
116 }
117 }
118 if (nowb!=0)
119 {
120 if (nowb>=a-nowa)
121 {
122 aa=a;
123 bb=nowb+nowa-a;
124 if (!flag[aa][bb])
125 {
126 tail++;
127 q[tail].a=aa;
128 q[tail].b=bb;
129 q[tail].d=3;
130 q[tail].d1=2;
131 wh[tail]=head;
132 flag[aa][bb]=1;
133 }
134 }
135 else if (nowb<a-nowa)
136 {
137 bb=0;
138 aa=nowa+nowb;
139 if (!flag[aa][bb])
140 {
141 tail++;
142 q[tail].a=aa;
143 q[tail].b=bb;
144 q[tail].d=3;
145 q[tail].d1=2;
146 wh[tail]=head;
147 flag[aa][bb]=1;
148 }
149 }
150 }
151 }
152 return -1;
153}
154void print(int k1)
155{
156 int i;
157 i=k1;
158 if (k1==-1)
159 {
160 printf("impossible\n");
161 return;
162 }
163 num=0;
164 while(i!=1)
165 {
166 num++;
167 num0[num]=i;
168 i=wh[i];
169 }
170 printf("%d\n",num);
171 for(i=num; i>=1; i--)
172 if (q[num0[i]].d==1)
173 {
174 printf("FILL(%d)\n",q[num0[i]].d1);
175 }
176 else if(q[num0[i]].d==2)
177 {
178 printf("DROP(%d)\n",q[num0[i]].d1);
179 }
180 else if(q[num0[i]].d==3)
181 {
182 if (q[num0[i]].d1==1)
183 {
184 printf("POUR(1,2)\n");
185 }
186 else if(q[num0[i]].d1==2)
187 {
188 printf("POUR(2,1)\n");
189 }
190 }
191}
192int main()
193{
194 scanf("%d%d%d",&a,&b,&c);
195 memset(flag,0,sizeof(flag));
196 if ((c>a)&&(c>b))
197 {
198 k=-1;
199 print(k);
200 }
201 else
202 {
203 k=bfs();
204 print(k);
205 }
206 return 0;
207}
208