poj 1077 Eight(经典八数码问题:bfs/Dbfs)
2016-06-03 16:08
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poj 1077 Eight(经典八数码问题:bDfs/Dbfs)
总时间限制: 5000ms 内存限制: 65536kB
Special Judge
描述
The 15-puzzle has been around for over 100 years; even if you don’t know it by that name, you’ve seen it. It is constructed with 15 sliding tiles, each with a number from 1 to 15 on it, and all packed into a 4 by 4 frame with one tile missing. Let’s call the missing tile ‘x’; the object of the puzzle is to arrange the tiles so that they are ordered as:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 x
where the only legal operation is to exchange ‘x’ with one of the tiles with which it shares an edge. As an example, the following sequence of moves solves a slightly scrambled puzzle:
The letters in the previous row indicate which neighbor of the ‘x’ tile is swapped with the ‘x’ tile at each step; legal values are ‘r’,’l’,’u’ and ‘d’, for right, left, up, and down, respectively.
Not all puzzles can be solved; in 1870, a man named Sam Loyd was famous for distributing an unsolvable version of the puzzle, and
frustrating many people. In fact, all you have to do to make a regular puzzle into an unsolvable one is to swap two tiles (not counting the missing ‘x’ tile, of course).
In this problem, you will write a program for solving the less well-known 8-puzzle, composed of tiles on a three by three
arrangement.
输入
You will receive a description of a configuration of the 8 puzzle. The description is just a list of the tiles in their initial positions, with the rows listed from top to bottom, and the tiles listed from left to right within a row, where the tiles are represented by numbers 1 to 8, plus ‘x’. For example, this puzzle
1 2 3
x 4 6
7 5 8
is described by this list:
1 2 3 x 4 6 7 5 8
输出
You will print to standard output either the word “unsolvable”, if the puzzle has no solution, or a string consisting entirely of the letters ‘r’, ‘l’, ‘u’ and ‘d’ that describes a series of moves that produce a solution. The string should include no spaces and start at the beginning of the line.
样例输入
2 3 4 1 5 x 7 6 8
样例输出
ullddrurdllurdruldr
来源
South Central USA 1998
本题是经典八数码问题,这一次以这道题为契机学习了Dbfs,现在已经学会了用bfs(version 1)和Dbfs(version 2)实现了。注意本题是special judge的,这样才不必考虑合理的搜索顺保证字典序,否则Dbfs无法安排搜索顺序…
注意一个编码小技巧Cantor launched,用来给全排列编码而且很节省空间。以后要用了推一推还是可以记起来的。
下面是代码(含测试部分)和评测结果,可以发现Dbfs快于bfs
version 1
version 2
总时间限制: 5000ms 内存限制: 65536kB
Special Judge
描述
The 15-puzzle has been around for over 100 years; even if you don’t know it by that name, you’ve seen it. It is constructed with 15 sliding tiles, each with a number from 1 to 15 on it, and all packed into a 4 by 4 frame with one tile missing. Let’s call the missing tile ‘x’; the object of the puzzle is to arrange the tiles so that they are ordered as:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 x
where the only legal operation is to exchange ‘x’ with one of the tiles with which it shares an edge. As an example, the following sequence of moves solves a slightly scrambled puzzle:
The letters in the previous row indicate which neighbor of the ‘x’ tile is swapped with the ‘x’ tile at each step; legal values are ‘r’,’l’,’u’ and ‘d’, for right, left, up, and down, respectively.
Not all puzzles can be solved; in 1870, a man named Sam Loyd was famous for distributing an unsolvable version of the puzzle, and
frustrating many people. In fact, all you have to do to make a regular puzzle into an unsolvable one is to swap two tiles (not counting the missing ‘x’ tile, of course).
In this problem, you will write a program for solving the less well-known 8-puzzle, composed of tiles on a three by three
arrangement.
输入
You will receive a description of a configuration of the 8 puzzle. The description is just a list of the tiles in their initial positions, with the rows listed from top to bottom, and the tiles listed from left to right within a row, where the tiles are represented by numbers 1 to 8, plus ‘x’. For example, this puzzle
1 2 3
x 4 6
7 5 8
is described by this list:
1 2 3 x 4 6 7 5 8
输出
You will print to standard output either the word “unsolvable”, if the puzzle has no solution, or a string consisting entirely of the letters ‘r’, ‘l’, ‘u’ and ‘d’ that describes a series of moves that produce a solution. The string should include no spaces and start at the beginning of the line.
样例输入
2 3 4 1 5 x 7 6 8
样例输出
ullddrurdllurdruldr
来源
South Central USA 1998
本题是经典八数码问题,这一次以这道题为契机学习了Dbfs,现在已经学会了用bfs(version 1)和Dbfs(version 2)实现了。注意本题是special judge的,这样才不必考虑合理的搜索顺保证字典序,否则Dbfs无法安排搜索顺序…
注意一个编码小技巧Cantor launched,用来给全排列编码而且很节省空间。以后要用了推一推还是可以记起来的。
下面是代码(含测试部分)和评测结果,可以发现Dbfs快于bfs
version 1
Accepted 9140kB 220ms 2398 B G++
#define TEST #undef TEST #define MAX_LEN 362880 #define TARGET_STATE 46233 #include<stdio.h> #include<iostream> using namespace std; const char operate_name[4]={'u','d','l','r'}; const int change[9][4]={{-1, 3,-1, 1},{-1, 4, 0, 2},{-1, 5, 1,-1}, { 0, 6,-1, 4},{ 1, 7, 3, 5},{ 2, 8, 4,-1}, { 3,-1,-1, 7},{ 4,-1, 6, 8},{ 5,-1, 7,-1}}; /* * 0 1 2 * 3 4 5 * 6 7 8 */ int queue[MAX_LEN][9]; int blank[MAX_LEN],father[MAX_LEN],operate[MAX_LEN]; int head,tail; char str_out[MAX_LEN]; int len; bool visit[MAX_LEN]; int cantor_launched(int a[]); void read_and_init(); void bfs(); void print(int last); void test(); int main() { #ifdef TEST freopen("input.txt","r",stdin); freopen("output.txt","w",stdout); #endif read_and_init(); bfs(); #ifdef TEST test(); #endif return 0; } int cantor_launched(int a[]) { int anti_num,frac=1,ans=0; for (int i=8;i>=0;i--) { anti_num=0; for (int j=i;j<9;j++) if (a[j]<a[i]) anti_num++; ans+=frac*anti_num; frac*=(9-i); } return ans; } void read_and_init() { char ch; for (int i=0;i<9;i++) { cin>>ch; if (ch=='X' || ch=='x') { blank[0]=i; queue[0][i]=0; } else queue[0][i]=ch-'0'; } head=0; tail=0; father[0]=0; operate[0]=0; return; } void bfs() { int new_blank; while (head<=tail) { if (cantor_launched(queue[head])==TARGET_STATE) { print(head); return; } for (int d=0;d<4;d++) { new_blank=change[blank[head]][d]; if (new_blank!=-1) { tail++; for (int i=0;i<9;i++) queue[tail][i]=queue[head][i]; queue[tail][new_blank]=0; queue[tail][blank[head]]=queue[head][new_blank]; if (visit[cantor_launched(queue[tail])]) tail--; else { visit[cantor_launched(queue[tail])]=true; blank[tail]=new_blank; father[tail]=head; operate[tail]=d; } } } head++; } printf("unsolvable\n"); return; } void print(int last) { while (father[last]!=last) { str_out[len++]=operate_name[operate[last]]; #ifdef TEST printf("\n"); for (int t=0;t<9;t++) printf("%d%c",queue[last][t],(t+1)%3?' ':'\n'); #endif last=father[last]; } for (int i=len-1;i>=0;i--) printf("%c",str_out[i]); printf("\n"); return; } void test() { for (int i=0;i<tail;i++) { for (int t=0;t<9;t++) printf("%d%c",queue[i][t],(t+1)%3?' ':'\n'); printf("father=%d,operate=%d\n\n",father[i],operate[i]); } return; }
version 2
Accepted 1928kB 0ms 4690 B G++
#define TEST #undef TEST #define MAX_LEN 362880 #define TARGET_STATE 46233 #include<stdio.h> #include<iostream> using namespace std; const char operate_name[4]={'u','d','l','r'}; const int change[9][4]={{-1, 3,-1, 1},{-1, 4, 0, 2},{-1, 5, 1,-1}, { 0, 6,-1, 4},{ 1, 7, 3, 5},{ 2, 8, 4,-1}, { 3,-1,-1, 7},{ 4,-1, 6, 8},{ 5,-1, 7,-1}}; const char operate_rname[4]={'d','u','r','l'}; const int rchange[9][4]={{ 3,-1, 1,-1},{ 4,-1, 2, 0},{ 5,-1,-1, 1}, { 6, 0, 4,-1},{ 7, 1, 5, 3},{ 8, 2,-1, 4}, {-1, 3, 7,-1},{-1, 4, 8, 6},{-1, 5,-1, 7}}; /* * 0 1 2 * 3 4 5 * 6 7 8 */ int queue[MAX_LEN+1][9]; int blank[MAX_LEN+1],father[MAX_LEN+1],operate[MAX_LEN+1]; int head,tail,rhead,rtail; char str_out[MAX_LEN]; int len; bool visit[MAX_LEN],rvisit[MAX_LEN]; int cantor_launched(int a[]); void read_and_init(); void dbfs(); void print(int last); void rprint(int last); void test() { printf("\n___________TEST___________\n"); for (int i=0;i<tail;i++) { printf("\n"); for (int t=0;t<9;t++) printf("%d%c",queue[i][t],(t+1)%3?' ':'\n'); printf("father=%5d,operate=%5d\n",father[i],operate[i]); } for (int i=rtail;i<=MAX_LEN;i++) { printf("\n"); for (int t=0;t<9;t++) printf("%d%c",queue[i][t],(t+1)%3?' ':'\n'); printf("father=%5d,operate=%5d\n",father[i],operate[i]); } return; } int main() { #ifdef TEST freopen("input.txt","r",stdin); freopen("output.txt","w",stdout); #endif read_and_init(); dbfs(); #ifdef TEST test(); #endif return 0; } int cantor_launched(int a[]) { int anti_num,frac=1,ans=0; for (int i=8;i>=0;i--) { anti_num=0; for (int j=i;j<9;j++) if (a[j]<a[i]) anti_num++; ans+=frac*anti_num; frac*=(9-i); } return ans; } void read_and_init() { char ch; for (int i=0;i<9;i++) { cin>>ch; if (ch=='X' || ch=='x') { blank[0]=i; queue[0][i]=0; } else queue[0][i]=ch-'0'; } head=0; tail=0; father[0]=0; operate[0]=0; visit[cantor_launched(queue[0])]=true; for (int i=0;i<8;i++) queue[MAX_LEN][i]=i+1; queue[MAX_LEN][8]=0; rhead=MAX_LEN; rtail=MAX_LEN; father[MAX_LEN]=MAX_LEN; operate[MAX_LEN]=0; blank[MAX_LEN]=8; rvisit[TARGET_STATE]=true; return; } void dbfs() { int new_blank; while (head<=tail || rhead>=rtail) { if (head<=tail){ if (rvisit[cantor_launched(queue[head])]) { print(head); for (int j=rtail;j<=rhead;j++) if (cantor_launched(queue[head])==cantor_launched(queue[j])) { rprint(j); break; } return; } for (int d=0;d<4;d++) { new_blank=change[blank[head]][d]; if (new_blank!=-1) { tail++; for (int i=0;i<9;i++) queue[tail][i]=queue[head][i]; queue[tail][new_blank]=0; queue[tail][blank[head]]=queue[head][new_blank]; if (visit[cantor_launched(queue[tail])]) tail--; else { visit[cantor_launched(queue[tail])]=true; blank[tail]=new_blank; father[tail]=head; operate[tail]=d; } } } head++;} if (rhead>=rtail){ if (visit[cantor_launched(queue[rhead])]) { for (int j=head;j<=tail;j++) if (cantor_launched(queue[rhead])==cantor_launched(queue[j])) { print(j); break; } rprint(rhead); return; } for (int d=0;d<4;d++) { new_blank=rchange[blank[rhead]][d]; if (new_blank!=-1) { rtail--; for (int i=0;i<9;i++) queue[rtail][i]=queue[rhead][i]; queue[rtail][new_blank]=0; queue[rtail][blank[rhead]]=queue[rhead][new_blank]; if (rvisit[cantor_launched(queue[rtail])]) rtail++; else { rvisit[cantor_launched(queue[rtail])]=true; blank[rtail]=new_blank; father[rtail]=rhead; operate[rtail]=d; } } } rhead--;} } printf("unsolvable\n"); return; } void print(int last) { #ifdef TEST printf("Steps pointers move:\nhead=%d,tail=%d,rhead=%d,rtail=%d\n", head,tail,MAX_LEN-rhead,MAX_LEN-rtail); printf("%d\n",last); #endif while (father[last]!=last) { str_out[len++]=operate_name[operate[last]]; last=father[last]; #ifdef TEST printf("\n"); for (int t=0;t<9;t++) printf("%d%c",queue[last][t],(t+1)%3?' ':'\n'); printf("father=%5d,operate=%5d\n",father[last],operate[last]); #endif } for (int i=len-1;i>=0;i--) printf("%c",str_out[i]); #ifdef TEST printf("\n"); #endif return; } void rprint(int last) { #ifdef TEST printf("%d\n",last); #endif len=0; while (father[last]!=last) { str_out[len++]=operate_name[operate[last]]; #ifdef TEST printf("\n"); for (int t=0;t<9;t++) printf("%d%c",queue[last][t],(t+1)%3?' ':'\n'); printf("father=%5d,operate=%5d\n",father[last],operate[last]); #endif last=father[last]; } str_out[len]=0; printf("%s\n",str_out); return; }
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