这道题目最开始做的时候wa+TLE。后面知道需要状态压缩，最近A掉。并且练习一下各种搜索算法。
1. 逆向BFS+康拓展开。
1 #include <iostream> 2 #include <queue> 3 #include <cstring> 4 #include <string> 5 #include <cstdio> 6 using namespace std; 7 8 #define N 9 9 #define MAXNUM 362880 10 11 typedef struct { 12 char map[N]; 13 int xpos; 14 string path; 15 } node_st; 16 17 int fact[N] = {1,1,2,6,24,120,720,5040,40320}; 18 int direct[4][2] = {{-1,0},{1,0},{0,-1},{0,1}}; 19 char rmove[4] = {'d','u','r','l'}; 20 char visit[MAXNUM]; 21 string rpath[MAXNUM]; 22 23 int cantor(node_st node) { 24 int i, j, cnt, val = 0; 25 26 for (i=0; i<N; ++i) { 27 cnt = 0; 28 for (j=i+1; j<N; ++j) { 29 if (node.map[j] < node.map[i]) 30 ++cnt; 31 } 32 val += fact[N-1-i]*cnt; 33 } 34 35 return val; 36 } 37 38 void bfs() { 39 queue<node_st> que; 40 node_st node; 41 int i, x, y, t1, t2, can; 42 43 for (i=1; i<N; ++i) 44 node.map[i-1] = '0' + i; 45 node.map[N-1] = 'x'; 46 node.xpos = N-1; 47 memset(visit, 0, sizeof(visit)); 48 visit[0] = 1; 49 que.push(node); 50 51 while ( !que.empty() ) { 52 node = que.front(); 53 que.pop(); 54 t1 = node.xpos / 3; 55 t2 = node.xpos % 3; 56 for (i=0; i<4; ++i) { 57 x = t1 + direct[i][0]; 58 y = t2 + direct[i][1]; 59 if (x<0 || x>=3 || y<0 || y>=3) 60 continue; 61 node_st tmp(node); 62 tmp.xpos = x*3+y; 63 tmp.map[node.xpos] = node.map[tmp.xpos]; 64 tmp.map[tmp.xpos] = 'x'; 65 can = cantor(tmp); 66 if ( !visit[can] ) { 67 visit[can] = 1; 68 tmp.path += rmove[i]; 69 rpath[can] = tmp.path; 70 que.push(tmp); 71 } 72 } 73 } 74 } 75 76 int main() { 77 node_st node; 78 int i, can; 79 80 bfs(); 81 82 while (scanf("%c%*c", &node.map[0]) != EOF) { 83 for (i=1; i<N; ++i) 84 scanf("%c%*c", &node.map[i]); 85 can = cantor(node); 86 if ( visit[can] ) { 87 for (i=rpath[can].size()-1; i>=0; --i) 88 printf("%c", rpath[can][i]); 89 printf(" "); 90 } else { 91 printf("unsolvable "); 92 } 93 } 94 95 return 0; 96 }
2. 双端BFS。注意逆序奇偶性相同才有解。
1 #include <iostream> 2 #include <queue> 3 #include <string> 4 #include <algorithm> 5 #include <cstring> 6 #include <cstdio> 7 using namespace std; 8 9 #define N 9 10 #define MAXNUM 362880 11 12 typedef struct { 13 char map[N]; 14 int xpos; 15 string path; 16 } node_st; 17 18 int fact[N]={1,1,2,6,24,120,720,5040,40320}; 19 int direct[4][2] = {{-1,0},{1,0},{0,-1},{0,1}}; 20 char move[4] = {'u','d','l','r'}; 21 char rmove[4] = {'d','u','r','l'}; 22 char visit[MAXNUM]; 23 bool flag; 24 string paths[MAXNUM], fpath; 25 queue<node_st> que, rque; 26 27 28 inline int cantor(node_st node) { 29 int i, j, cnt, val = 0; 30 31 for (i=0; i<N; ++i) { 32 cnt = 0; 33 for (j=i+1; j<N; ++j) 34 if (node.map[j] < node.map[i]) 35 ++cnt; 36 val += fact[N-1-i]*cnt; 37 } 38 39 return val; 40 } 41 42 inline bool invNum(node_st node) { 43 int i, j, cnt = 0; 44 45 for (i=0; i<N; ++i) { 46 if (node.map[i] == 'x') 47 continue; 48 for (j=i-1; j>=0; --j) 49 if (node.map[j]!='x' && node.map[j]>node.map[i]) 50 ++cnt; 51 } 52 53 return cnt&1; 54 } 55 56 void bfs() { 57 int x, y, can; 58 node_st node = que.front(); 59 que.pop(); 60 61 for (int i=0; i<4; ++i) { 62 x = node.xpos/3 + direct[i][0]; 63 y = node.xpos%3 + direct[i][1]; 64 if (x<0 || x>=3 || y<0 || y>=3) 65 continue; 66 node_st p(node); 67 p.map[p.xpos] = node.map[x*3+y]; 68 p.xpos = x*3+y; 69 p.map[p.xpos] = 'x'; 70 can = cantor(p); 71 if (visit[can] == 1) 72 continue; 73 p.path += move[i]; 74 if (visit[can] == -1) { 75 flag = false; 76 reverse(paths[can].begin(), paths[can].end()); 77 fpath = p.path + paths[can]; 78 return ; 79 } 80 visit[can] = 1; 81 paths[can] = p.path; 82 que.push(p); 83 } 84 } 85 86 void rbfs() { 87 int x, y, can; 88 node_st node = rque.front(); 89 rque.pop(); 90 91 for (int i=0; i<4; ++i) { 92 x = node.xpos/3 + direct[i][0]; 93 y = node.xpos%3 + direct[i][1]; 94 if (x<0 || x>=3 || y<0 || y>=3) 95 continue; 96 node_st p(node); 97 p.map[p.xpos] = node.map[x*3+y]; 98 p.xpos = x*3+y; 99 p.map[p.xpos] = 'x'; 100 can = cantor(p); 101 if (visit[can] == -1) 102 continue; 103 p.path += rmove[i]; 104 if (visit[can] == 1) { 105 flag = false; 106 reverse(p.path.begin(), p.path.end()); 107 fpath = paths[can] + p.path; 108 } 109 visit[can] = -1; 110 paths[can] = p.path; 111 rque.push(p); 112 } 113 } 114 115 int main() { 116 node_st node, enode; 117 int i, n, can; 118 119 for (i=1; i<N; ++i) 120 enode.map[i-1] = i + '0'; 121 enode.map[N-1] = 'x'; 122 enode.xpos = N-1; 123 124 while (scanf("%c%*c", &node.map[0]) != EOF) { 125 for (i=0; i<N; ++i) { 126 if (i) 127 scanf("%c%*c", &node.map[i]); 128 if (node.map[i] == 'x') 129 node.xpos = i; 130 } 131 if ( invNum(node) ) { 132 printf("unsolvable "); 133 continue; 134 } 135 can = cantor(node); 136 if ( can == 0 ) { 137 printf(" "); 138 continue; 139 } 140 while ( !que.empty() ) 141 que.pop(); 142 while ( !rque.empty() ) 143 rque.pop(); 144 memset(visit, 0, sizeof(visit)); 145 flag = true; 146 rque.push(enode); 147 visit[0] = -1; 148 que.push(node); 149 visit[can] = 1; 150 while (flag) { 151 n = que.size(); 152 while (flag && n--) 153 bfs(); 154 n = rque.size(); 155 while (flag && n--) 156 rbfs(); 157 } 158 cout <<fpath<<endl; 159 } 160 161 return 0; 162 }
3. A-star h函数为曼哈顿距离，必须用内联才不会TLE。
1 #include <iostream> 2 #include <queue> 3 #include <stack> 4 #include <string> 5 #include <algorithm> 6 #include <cstring> 7 #include <cstdio> 8 using namespace std; 9 10 #define N 9 11 #define MAXN 362880 12 //#define myabs(x) ((x)>0) ? (x):(-(x)) 13 14 typedef struct node_st { 15 char map[N]; 16 int xpos; 17 int s, p; 18 friend bool operator < (node_st a, node_st b) { 19 return a.p > b.p; 20 } 21 } node_st; 22 23 char visit[MAXN]; 24 int fact[N] = {1,1,2,6,24,120,720,5040,40320}; 25 int direct[4][2] = {{-1,0},{1,0},{0,-1},{0,1}}; 26 char move[4] = {'u', 'd', 'l', 'r'}; 27 int pre[MAXN], bcan; 28 char op[MAXN]; 29 node_st beg; 30 31 inline int cantor(node_st node) { 32 int i, j, cnt, val = 0; 33 34 for (i=0; i<N; ++i) { 35 cnt = 0; 36 for (j=i+1; j<N; ++j) 37 if (node.map[j] < node.map[i]) 38 ++cnt; 39 val += fact[N-1-i]*cnt; 40 } 41 42 return val; 43 } 44 45 inline myabs(int x) { 46 return (x>0) ? x:-x; 47 } 48 49 inline int h(node_st node) { 50 int tmp, val = 0; 51 52 for (int i=0; i<N; ++i) { 53 if (node.map[i] == 'x') continue; 54 tmp = node.map[i] - '1'; 55 val += myabs(tmp/3-i/3) + myabs(tmp%3-i%3); 56 } 57 58 return val; 59 } 60 61 inline bool invNum(node_st node) { 62 int i, j, cnt = 0; 63 64 for (i=0; i<N; ++i) { 65 if (node.map[i] == 'x') continue; 66 for (j=i-1; j>=0; --j) 67 if (node.map[j]!='x' && node.map[j]>node.map[i]) 68 ++cnt; 69 } 70 71 return cnt&1; 72 } 73 74 void bfs() { 75 priority_queue<node_st> que; 76 node_st node; 77 int i, t1, t2, x, y, can, pcan; 78 79 que.push(beg); 80 memset(visit, 0, sizeof(visit)); 81 visit[bcan] = 1; 82 83 while ( !que.empty() ) { 84 node = que.top(); 85 que.pop(); 86 t1 = node.xpos / 3; 87 t2 = node.xpos % 3; 88 pcan = cantor(node); 89 for (i=0; i<4; ++i) { 90 x = t1 + direct[i][0]; 91 y = t2 + direct[i][1]; 92 if (x<0 || x>=3 || y<0 || y>=3) 93 continue; 94 node_st p(node); 95 p.map[node.xpos] = p.map[x*3+y]; 96 p.xpos = x*3+y; 97 p.map[p.xpos] = 'x'; 98 can = cantor(p); 99 if (visit[can]) 100 continue; 101 visit[can] = 1; 102 ++p.s; 103 p.p = p.s + h(p); 104 pre[can] = pcan; 105 op[can] = move[i]; 106 if (can == 0) 107 return ; 108 que.push(p); 109 } 110 } 111 } 112 113 int main() { 114 int i, k; 115 stack<char> st; 116 117 while (scanf("%c%*c", &beg.map[0]) != EOF) { 118 for (i=0; i<N; ++i) { 119 if (i) 120 scanf("%c%*c", &beg.map[i]); 121 if (beg.map[i] == 'x') 122 beg.xpos = i; 123 } 124 beg.s = 0; 125 beg.p = h(beg); 126 if ( invNum(beg) ) { 127 printf("unsolvable "); 128 continue; 129 } 130 bcan = cantor(beg); 131 if (bcan == 0) { 132 printf(" "); 133 continue; 134 } 135 bfs(); 136 k = 0; 137 while (k != bcan) { 138 st.push(op[k]); 139 k = pre[k]; 140 } 141 142 while ( !st.empty() ) { 143 printf("%c", st.top()); 144 st.pop(); 145 } 146 printf(" "); 147 } 148 return 0; 149 }
4. IDA*，该算法其实是迭代dfs。H函数还是曼哈顿距离。
1 #include <iostream> 2 #include <string> 3 #include <cstdio> 4 #include <cstring> 5 using namespace std; 6 7 #define N 9 8 #define MAXN 362880 9 10 typedef struct { 11 char map[N]; 12 int xpos; 13 } node_st; 14 15 int fact[N] = {1,1,2,6,24,120,720,5040,40320}; 16 int direct[4][2] = {{-1,0},{1,0},{0,-1},{0,1}}; 17 char move[4] = {'u', 'd', 'l', 'r'}; 18 char visit[MAXN]; 19 char op[MAXN]; 20 int deep; 21 bool flag; 22 node_st beg; 23 24 inline int cantor(node_st node) { 25 int i, j, cnt, val = 0; 26 27 for (i=0; i<N; ++i) { 28 cnt = 0; 29 for (j=i+1; j<N; ++j) 30 if (node.map[j] < node.map[i]) 31 ++cnt; 32 val += fact[N-1-i]*cnt; 33 } 34 35 return val; 36 } 37 38 inline bool invNum(node_st node) { 39 int i, j, cnt = 0; 40 41 for (i=0; i<N; ++i) { 42 if (node.map[i] == 'x') continue; 43 for (j=i-1; j>=0; --j) 44 if (node.map[j]!='x' && node.map[j]>node.map[i]) 45 ++cnt; 46 } 47 48 return cnt&1; 49 } 50 51 inline int myabs(int x) { 52 return (x<0) ? -x:x; 53 } 54 55 inline int h(node_st node) { 56 int i, tmp, val = 0; 57 58 for (i=0; i<N; ++i) { 59 if (node.map[i] == 'x') continue; 60 tmp = node.map[i] - '1'; 61 val += myabs(tmp/3-i/3) + myabs(tmp%3-i%3); 62 } 63 64 return val; 65 } 66 67 void dfs(int d) { 68 int i, t1, t2, x, y, can; 69 70 t1 = h(beg); 71 if (t1 == 0) { 72 flag = false; 73 return ; 74 } 75 if (t1+d > deep) 76 return ; 77 t1 = beg.xpos / 3; 78 t2 = beg.xpos % 3; 79 for (i=0; i<4; ++i) { 80 x = t1 + direct[i][0]; 81 y = t2 + direct[i][1]; 82 if (x<0 || x>=3 || y<0 || y>=3) 83 continue; 84 node_st p(beg), org(beg); 85 p.map[p.xpos] = p.map[x*3+y]; 86 p.xpos = x*3+y; 87 p.map[p.xpos] = 'x'; 88 can = cantor(p); 89 if (visit[can]) continue; 90 visit[can] = 1; 91 op[d] = move[i]; 92 beg = p; 93 dfs(d+1); 94 if ( !flag ) 95 return ; 96 beg = org; 97 visit[can] = 0; 98 } 99 } 100 101 int main() { 102 int i; 103 104 while (scanf("%c%*c", &beg.map[0]) != EOF) { 105 for (i=0; i<N; ++i) { 106 if (i) 107 scanf("%c%*c", &beg.map[i]); 108 if (beg.map[i] == 'x') 109 beg.xpos = i; 110 } 111 if ( invNum(beg) ) { 112 printf("unsolvable "); 113 continue; 114 } 115 memset(visit, 0, sizeof(visit)); 116 memset(op, 0, sizeof(op)); 117 i = cantor(beg); 118 visit[i] = 1; 119 deep = h(beg); 120 flag = true; 121 while (flag) { 122 ++deep; 123 dfs(0); 124 //printf("%d ", deep); 125 } 126 127 i = 0; 128 while ( op[i] ) 129 printf("%c", op[i++]); 130 printf(" "); 131 } 132 133 return 0; 134 }