1 /*
2 题意：八数码问题，给出3*3的矩阵含1~8以及x，给出一个符合的解使得移动后的矩阵的顺序为1~8，最后为x
3
4 题解：BFS
5 需要用到康托展开来表示状态，不然数组无法完全表示所有状态，这样BFS就无法判断找不到解的情况(status
6 的0ms,0KB究竟是怎么做到的，简直不能想象=。=)
7 */
8 #include <cstdio>
9 #include <cstring>
10 #include <queue>
11 #include <stack>
12 #include <algorithm>
13
14 using namespace std;
15
16 int fac[] = {1,1,2,6,24,120,720,5040,40320,362880}; //i的阶乘为fac[i]
17 /* 康托展开.
18 {1...n}的全排列由小到大有序，s[]为第几个数 */
19 int KT(int n, int s[])
20 {
21 int i, j, t, sum;
22 sum = 0;
23 for (i=0; i<n; i++)
24 {
25 t = 0;
26 for (j=i+1; j<n; j++)
27 if (s[j] < s[i])
28 t++;
29 sum += t*fac[n-i-1];
30 }
31 return sum+1;
32 }
33
34 /* 康托展开的逆运算.
35 {1...n}的全排列，中的第k个数为s[] */
36 void invKT(int n, int k, int s[])
37 {
38 int i, j, t, vst[10]={0};
39 k--;
40 for (i=0; i<n; i++)
41 {
42 t = k/fac[n-i-1];
43 for (j=1; j<=n; j++)
44 if (!vst[j])
45 {
46 if (t == 0) break;
47 t--;
48 }
49 s[i] = j;
50 vst[j] = 1;
51 k %= fac[n-i-1];
52 }
53 }
54
55 int dir[4][2]={0,1,0,-1,1,0,-1,0};
56
57 int vis[370000];
58 char path[370000];
59 char ansstr[370000];
60
61 struct node
62 {
63 int state,pdir;
64 int x,y;
65 }Q[370000];
66 int front,tail;
67
68 int finish_state;
69
70 int bfs(node start)
71 {
72 memset(vis,-1,sizeof(vis));
73 vis[start.state] = 100000000;
74 front = tail = 0;
75 Q[tail++] = start;
76 while (front < tail)
77 {
78 node now = Q[front++];
79 if (now.state == finish_state)
80 return finish_state;
81 for(int i=0; i<4; i++)
82 {
83 node tmp = now;
84 if (tmp.pdir != i)
85 {
86 int s[10];
87 invKT(9,tmp.state,s);
88
89 int x = now.x, y = now.y;
90 int nx = x+dir[i][0],ny = y+dir[i][1];
91 if (0<=nx&&nx<=2 && 0<=ny&&ny<=2)
92 {
93 swap(s[x*3+y],s[nx*3+ny]);
94 tmp.pdir = i;
95 tmp.x = nx;
96 tmp.y = ny;
97 tmp.state = KT(9,s);
98 if (vis[tmp.state] != -1)
99 continue;
100 Q[tail++] = tmp;
101 vis[tmp.state] = now.state;
102 char c;
103 switch(i)
104 {
105 case 0:c = ‘r‘; break;
106 case 1:c = ‘l‘; break;
107 case 2:c = ‘d‘; break;
108 case 3:c = ‘u‘; break;
109 }
110 path[tmp.state] = c;
111 }
112 }
113 }
114 }
115 return -1;
116 }
117
118 int main(void)
119 {
120 char str[50],tstr[3];
121 int tmps[9]={1,2,3,4,5,6,7,8,9};
122 while (~scanf("%s",tstr))
123 {
124 str[0] = tstr[0];
125 for(int i=1; i<9; i++)
126 {
127 scanf("%s",tstr);
128 str[i*2] = tstr[0];
129 }
130 int s[9];
131 int x;
132 for(int i=0; i<9; i++)
133 {
134 if (‘1‘<=str[i*2] && str[i*2]<=‘9‘)
135 s[i] = str[i*2]-‘0‘;
136 else
137 {
138 s[i] = 9;
139 x = i;
140 }
141 }
142 // 输入略恶心，最后用s[]记录矩阵
143 node start;
144 start.pdir = -1;
145 start.x = x/3;
146 start.y = x%3;
147 start.state = KT(9,s);
148
149 finish_state = KT(9,tmps);
150 int ans = bfs(start);
151
152 if (ans < 0)
153 {
154 printf("unsolvable\n");
155 }
156 else
157 {
158 int sum = 0;
159 while (vis[finish_state] != -1)
160 {
161 if (vis[finish_state] == 100000000)
162 break;
163 ansstr[sum++] = path[finish_state];
164 finish_state = vis[finish_state];
165 }
166 ansstr[sum] = 0;
167
168 for(int i = sum-1; i>=0; i--)
169 printf("%c",ansstr[i]);
170 printf("\n");
171 }
172 }
173 return 0;
174 }
poj 1077 八数码(BFS+康托展开),布布扣,bubuko.com
时间： 2024-10-21 14:52:59