题目传送门
1 /*
2 BFS：六种情况讨论一下，BFS轻松解决
3 起初我看有人用DFS，我写了一遍，TLE。。还是用BFS，结果特判时出错，逗了好长时间
4 看别人的代码简直是受罪，还好自己终于发现自己代码的小错误：)
5 */
6 /************************************************
7 Author :Running_Time
8 Created Time :2015-8-3 14:17:24
9 File Name :POJ_3414_BFS.cpp
10 **************************************************/
11
12 #include <cstdio>
13 #include <algorithm>
14 #include <iostream>
15 #include <sstream>
16 #include <cstring>
17 #include <cmath>
18 #include <string>
19 #include <vector>
20 #include <queue>
21 #include <deque>
22 #include <stack>
23 #include <list>
24 #include <map>
25 #include <set>
26 #include <bitset>
27 #include <cstdlib>
28 #include <ctime>
29 using namespace std;
30
31 #define lson l, mid, rt << 1
32 #define rson mid + 1, r, rt << 1 | 1
33 typedef long long ll;
34 const int MAXN = 1e4 + 10;
35 const int INF = 0x3f3f3f3f;
36 const int MOD = 1e9 + 7;
37 struct Point {
38 int a, b, step;
39 int op[MAXN];
40 };
41 bool vis[110][110];
42 int A, B, C;
43 int ans;
44
45 void BFS(void) {
46 memset (vis, false, sizeof (vis));
47 queue<Point> Q; Q.push ((Point) {0, 0, 0});
48 while (!Q.empty ()) {
49 Point p = Q.front (); Q.pop ();
50 if (p.a == C || p.b == C) {
51 printf ("%d
", p.step);
52 for (int i=1; i<=p.step; ++i) {
53 if (p.op[i] == 1) puts ("FILL(1)");
54 else if (p.op[i] == 2) puts ("FILL(2)");
55 else if (p.op[i] == 3) puts ("DROP(1)");
56 else if (p.op[i] == 4) puts ("DROP(2)");
57 else if (p.op[i] == 5) puts ("POUR(1,2)");
58 else if (p.op[i] == 6) puts ("POUR(2,1)");
59 }
60 return ;
61 }
62 Point tmp;
63 if (p.a < A && !vis[A][p.b]) {
64 vis[A][p.b] = true;
65 tmp = p; tmp.a = A; tmp.op[++tmp.step] = 1; //FILL1
66 Q.push (tmp);
67 }
68 if (p.b < B && !vis[p.a][B]) {
69 vis[p.a][B] = true;
70 tmp = p; tmp.b = B; tmp.op[++tmp.step] = 2; //FILL2
71 Q.push (tmp);
72 }
73 if (p.a > 0 && !vis[0][p.b]) {
74 vis[0][p.b] = true;
75 tmp = p; tmp.a = 0; tmp.op[++tmp.step] = 3; //DROP1
76 Q.push (tmp);
77 }
78 if (p.b > 0 && !vis[p.a][0]) {
79 vis[p.a][0] = true;
80 tmp = p; tmp.b = 0; tmp.op[++tmp.step] = 4; //DROP2
81 Q.push (tmp);
82 }
83 if (p.a > 0 && p.b < B) {
84 int t = min (p.a, B - p.b);
85 if (!vis[p.a-t][p.b+t]) {
86 vis[p.a-t][p.b+t] = true; //POUR1->2
87 tmp = p; tmp.a -= t; tmp.b += t; tmp.op[++tmp.step] = 5;
88 Q.push (tmp);
89 }
90 }
91 if (p.b > 0 && p.a < A) {
92 int t = min (p.b, A - p.a);
93 if (!vis[p.a+t][p.b-t]) {
94 vis[p.a+t][p.b-t] = true; //POUR2->1
95 tmp = p; tmp.a += t; tmp.b -= t; tmp.op[++tmp.step] = 6;
96 Q.push (tmp);
97 }
98 }
99 }
100
101 puts ("impossible");
102 }
103
104 int main(void) { //POJ 3414 Pots
105 while (scanf ("%d%d%d", &A, &B, &C) == 3) {
106 BFS ();
107 }
108
109 return 0;
110 }