转自http://www.cnblogs.com/whatbeg/p/3728557.html
直接BFS肯定不行，复杂度太高。
先不考虑加加减操作，因为它们不涉及以为，很好处理。
因为一开始魔棒是在最左端，所以第i个位置被访问过了，则前面的一定都访问过。同时，我们可以直接通过和最后一位交换的方式访问最后一位。所以所有被访问的状态(标号)就只有1、12、123、1234、12345、123456、16、126、1236、12346，就10个。
所以状态记录就是dis[6][6][6][6][6][6][6][10]，前6个6 代表现在的第i位是原来的第j位，第7个6 代表魔杖现在在哪，10 代表有哪些位置被访问过了。
dis表示到当前状态的步数（最小）。
BFS出来后，枚举6位数的每一位和指针位置和10种状态，看此状态是否已访问，如果已访问，表示能够到达这个状态，然后看从这个状态到目标状态还需多少步（有可能达不到），然后看dis[状态]+步数是否小于最小步数，是则更新答案。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <cmath>
5 #include <algorithm>
6 #include <queue>
7 #define Mod 1000000007
8 #define INT 2147483647
9 using namespace std;
10 #define N 50007
11
12 struct node
13 {
14 int p[7];
15 int point,state;
16 int step;
17 }S;
18
19 void Popy(int *p1,int *p2)
20 {
21 for(int i=1;i<7;i++)
22 p1[i] = p2[i];
23 }
24
25 int dis[7][7][7][7][7][7][7][10];
26 int E[7],SS[7];
27
28 int min(int ka,int kb)
29 {
30 if(ka < kb)
31 return ka;
32 return kb;
33 }
34
35 int max(int ka,int kb)
36 {
37 if(ka > kb)
38 return ka;
39 return kb;
40 }
41
42 void BFS(node S)
43 {
44 queue<node> que;
45 memset(dis,-1,sizeof(dis));
46 que.push(S);
47 dis[S.p[1]][S.p[2]][S.p[3]][S.p[4]][S.p[5]][S.p[6]][1][0] = 0;
48 while(!que.empty())
49 {
50 node tmp = que.front();
51 que.pop();
52 for(int i=0;i<4;i++)
53 {
54 node now;
55 if(i == 0) //与左边交换
56 {
57 swap(tmp.p[1],tmp.p[tmp.point]);
58 Popy(now.p,tmp.p);
59 swap(tmp.p[1],tmp.p[tmp.point]);
60 now.point = tmp.point;
61 now.state = tmp.state;
62 now.step = tmp.step + 1;
63 if(dis[now.p[1]][now.p[2]][now.p[3]][now.p[4]][now.p[5]][now.p[6]][now.point][now.state] == -1 || (dis[now.p[1]][now.p[2]][now.p[3]][now.p[4]][now.p[5]][now.p[6]][now.point][now.state] != -1 && now.step < dis[now.p[1]][now.p[2]][now.p[3]][now.p[4]][now.p[5]][now.p[6]][now.point][now.state]))
64 {
65 dis[now.p[1]][now.p[2]][now.p[3]][now.p[4]][now.p[5]][now.p[6]][now.point][now.state] = now.step;
66 que.push(now);
67 }
68 }
69 else if(i == 1) //与右边交换
70 {
71 swap(tmp.p[6],tmp.p[tmp.point]);
72 Popy(now.p,tmp.p);
73 swap(tmp.p[6],tmp.p[tmp.point]);
74 now.point = tmp.point;
75 if(tmp.state <= 3)
76 now.state = tmp.state + 6;
77 else if(tmp.state == 4)
78 now.state = tmp.state + 1;
79 else
80 now.state = tmp.state;
81 now.step = tmp.step + 1;
82 if(dis[now.p[1]][now.p[2]][now.p[3]][now.p[4]][now.p[5]][now.p[6]][now.point][now.state] == -1 || (dis[now.p[1]][now.p[2]][now.p[3]][now.p[4]][now.p[5]][now.p[6]][now.point][now.state] != -1 && now.step < dis[now.p[1]][now.p[2]][now.p[3]][now.p[4]][now.p[5]][now.p[6]][now.point][now.state]))
83 {
84 dis[now.p[1]][now.p[2]][now.p[3]][now.p[4]][now.p[5]][now.p[6]][now.point][now.state] = now.step;
85 que.push(now);
86 }
87 }
88 else if(i == 2) //左移
89 {
90 Popy(now.p,tmp.p);
91 now.point = max(1,tmp.point-1);
92 now.state = tmp.state;
93 now.step = tmp.step + 1;
94 if(dis[now.p[1]][now.p[2]][now.p[3]][now.p[4]][now.p[5]][now.p[6]][now.point][now.state] == -1 || (dis[now.p[1]][now.p[2]][now.p[3]][now.p[4]][now.p[5]][now.p[6]][now.point][now.state] != -1 && now.step < dis[now.p[1]][now.p[2]][now.p[3]][now.p[4]][now.p[5]][now.p[6]][now.point][now.state]))
95 {
96 dis[now.p[1]][now.p[2]][now.p[3]][now.p[4]][now.p[5]][now.p[6]][now.point][now.state] = now.step;
97 que.push(now);
98 }
99 }
100 else if(i == 3) //右移
101 {
102 Popy(now.p,tmp.p);
103 now.point = min(6,tmp.point+1);
104 if(tmp.state < 5)
105 now.state = tmp.state+1;
106 else if(tmp.state == 5)
107 now.state = tmp.state;
108 else if(tmp.state < 9)
109 now.state = tmp.state+1;
110 else
111 now.state = tmp.state-4;
112 now.step = tmp.step + 1;
113 if(dis[now.p[1]][now.p[2]][now.p[3]][now.p[4]][now.p[5]][now.p[6]][now.point][now.state] == -1 || (dis[now.p[1]][now.p[2]][now.p[3]][now.p[4]][now.p[5]][now.p[6]][now.point][now.state] != -1 && now.step < dis[now.p[1]][now.p[2]][now.p[3]][now.p[4]][now.p[5]][now.p[6]][now.point][now.state]))
114 {
115 dis[now.p[1]][now.p[2]][now.p[3]][now.p[4]][now.p[5]][now.p[6]][now.point][now.state] = now.step;
116 que.push(now);
117 }
118 }
119 }
120 }
121 }
122
123 int main()
124 {
125 int a,b,i;
126 int h[7];
127 int pot,status;
128 while(scanf("%d%d",&a,&b)!=EOF)
129 {
130 S.p[1] = 1; //标号。初始都在自己的位置
131 S.p[2] = 2;
132 S.p[3] = 3;
133 S.p[4] = 4;
134 S.p[5] = 5;
135 S.p[6] = 6;
136 S.point = 1;
137 S.state = 0;
138 S.step = 0;
139 SS[1] = (a/100000)%10;
140 SS[2] = (a/10000)%10;
141 SS[3] = (a/1000)%10;
142 SS[4] = (a/100)%10;
143 SS[5] = (a/10)%10;
144 SS[6] = a%10;
145 E[1] = (b/100000)%10;
146 E[2] = (b/10000)%10;
147 E[3] = (b/1000)%10;
148 E[4] = (b/100)%10;
149 E[5] = (b/10)%10;
150 E[6] = b%10;
151 BFS(S);
152 int ans = Mod;
153 for(h[1]=1;h[1]<=6;h[1]++)
154 {
155 for(h[2]=1;h[2]<=6;h[2]++)
156 {
157 if(h[2] != h[1])
158 {
159 for(h[3]=1;h[3]<=6;h[3]++)
160 {
161 if(h[3] != h[2] && h[3] != h[1])
162 {
163 for(h[4]=1;h[4]<=6;h[4]++)
164 {
165 if(h[4] != h[1] && h[4] != h[2] && h[4] != h[3])
166 {
167 for(h[5]=1;h[5]<=6;h[5]++)
168 {
169 if(h[5] != h[1] && h[5] != h[2] && h[5] != h[3] && h[5] != h[4])
170 {
171 for(h[6]=1;h[6]<=6;h[6]++)
172 {
173 if(h[6] != h[1] && h[6] != h[2] && h[6] != h[3] && h[6] != h[4] && h[6] != h[5])
174 {
175 for(pot=1;pot<=6;pot++)
176 {
177 for(status=0;status<10;status++)
178 {
179 if(dis[h[1]][h[2]][h[3]][h[4]][h[5]][h[6]][pot][status] != -1)
180 {
181 int cnt = 0;
182 int t,r;
183 int flag = 1; //No. status
184 if(status <= 5) //1 1
185 { //2 12
186 for(t=1;t<=status+1;t++) //3 123
187 cnt += abs(E[t]-SS[h[t]]); //4 1234
188 for(t;t<=6;t++) //5 12345
189 if(E[t] != SS[h[t]]) //6 123456
190 { //7 16
191 flag = 0; //8 126
192 break; //9 1236
193 } //10 12346
194 if(!flag)
195 continue;
196 }
197 else if(status >= 6 && status <= 9)
198 {
199 for(r=1;r<=status-5;r++)
200 cnt += abs(E[r]-SS[h[r]]);
201 for(r;r<6;r++)
202 if(E[r] != SS[h[r]])
203 {
204 flag = 0;
205 break;
206 }
207 if(!flag)
208 continue;
209 cnt += abs(E[6]-SS[h[6]]);
210 }
211 ans = min(ans,dis[h[1]][h[2]][h[3]][h[4]][h[5]][h[6]][pot][status]+cnt);
212 }
213 }
214 }
215 }
216 }
217 }
218 }
219 }
220 }
221 }
222 }
223 }
224 }
225 }
226 printf("%d\n",ans);
227 }
228 return 0;
229 }
时间： 2024-10-02 04:40:44