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 ",ans); 227 } 228 return 0; 229 }