M × N Puzzle
| Time Limit: 4000MS | Memory Limit: 131072K | |
| Total Submissions: 4860 | Accepted: 1321 |
Description
The Eight Puzzle, among other sliding-tile puzzles, is one of the famous problems in artificial intelligence. Along with chess, tic-tac-toe and backgammon, it has been used to study search algorithms.
The Eight Puzzle can be generalized into an M × N Puzzle where at least one of M and N is odd. The puzzle is constructed with MN − 1 sliding tiles with each a number from 1 toMN − 1 on it packed into a M by N frame with one tile missing. For example, with M = 4 and N = 3, a puzzle may look like:
| 1 | 6 | 2 |
| 4 | 0 | 3 |
| 7 | 5 | 9 |
| 10 | 8 | 11 |
Let's call missing tile 0. The only legal operation is to exchange 0 and the tile with which it shares an edge. The goal of the puzzle is to find a sequence of legal operations that makes it look like:
| 1 | 2 | 3 |
| 4 | 5 | 6 |
| 7 | 8 | 9 |
| 10 | 11 | 0 |
The following steps solve the puzzle given above.
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START |
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DOWN |
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LEFT ⇒ |
|
UP |
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… |
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RIGHT |
|
UP |
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UP ⇒ |
|
LEFT |
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GOAL |
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Given an M × N puzzle, you are to determine whether it can be solved.
Input
The input consists of multiple test cases. Each test case starts with a line containing M and N (2 ≤ M, N ≤ 999). This line is followed by M lines containing N numbers each describing an M × N puzzle.
The input ends with a pair of zeroes which should not be processed.
Output
Output one line for each test case containing a single word YES if the puzzle can be solved and NO otherwise.
Sample Input
3 3 1 0 3 4 2 5 7 8 6 4 3 1 2 5 4 6 9 11 8 10 3 7 0 0 0
Sample Output
YES NO
1 #include<cstdio> 2 //#include<iostream> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 #include<vector> 7 //#include<queue> 8 //#include<set> 9 #define INF 0x3f3f3f3f 10 #define N 10000005 11 #define re register 12 #define Ii inline int 13 #define Il inline long long 14 #define Iv inline void 15 #define Ib inline bool 16 #define Id inline double 17 #define ll long long 18 #define Fill(a,b) memset(a,b,sizeof(a)) 19 #define R(a,b,c) for(register int a=b;a<=c;++a) 20 #define nR(a,b,c) for(register int a=b;a>=c;--a) 21 #define Min(a,b) ((a)<(b)?(a):(b)) 22 #define Max(a,b) ((a)>(b)?(a):(b)) 23 #define Cmin(a,b) ((a)=(a)<(b)?(a):(b)) 24 #define Cmax(a,b) ((a)=(a)>(b)?(a):(b)) 25 #define D_e(x) printf(" &__ %d __& ",x) 26 #define D_e_Line printf("----------------- ") 27 #define D_e_Matrix for(re int i=1;i<=n;++i){for(re int j=1;j<=m;++j)printf("%d ",g[i][j]);putchar(' ');} 28 using namespace std; 29 // The Code Below Is Bingoyes's Function Forest. 30 31 Ii read(){ 32 int s=0,f=1;char c; 33 for(c=getchar();c>'9'||c<'0';c=getchar())if(c=='-')f=-1; 34 while(c>='0'&&c<='9')s=s*10+(c^'0'),c=getchar(); 35 return s*f; 36 } 37 Iv print(ll x){ 38 if(x<0)putchar('-'),x=-x; 39 if(x>9)print(x/10); 40 putchar(x%10^'0'); 41 } 42 /* 43 Iv Floyd(){ 44 R(k,1,n) 45 R(i,1,n) 46 if(i!=k&&dis[i][k]!=INF) 47 R(j,1,n) 48 if(j!=k&&j!=i&&dis[k][j]!=INF) 49 Cmin(dis[i][j],dis[i][k]+dis[k][j]); 50 } 51 Iv Dijkstra(int st){ 52 priority_queue<int>q; 53 R(i,1,n)dis[i]=INF; 54 dis[st]=0,q.push((nod){st,0}); 55 while(!q.empty()){ 56 int u=q.top().x,w=q.top().w;q.pop(); 57 if(w!=dis[u])continue; 58 for(re int i=head[u];i;i=e[i].nxt){ 59 int v=e[i].pre; 60 if(dis[v]>dis[u]+e[i].w) 61 dis[v]=dis[u]+e[i].w,q.push((nod){v,dis[v]}); 62 } 63 } 64 } 65 Iv Count_Sort(int arr[]){ 66 int k=0; 67 R(i,1,n) 68 ++tot[arr[i]],Cmax(mx,a[i]); 69 R(j,0,mx) 70 while(tot[j]) 71 arr[++k]=j,--tot[j]; 72 } 73 Iv Merge_Sort(int arr[],int left,int right,int &sum){ 74 if(left>=right)return; 75 int mid=left+right>>1; 76 Merge_Sort(arr,left,mid,sum),Merge_Sort(arr,mid+1,right,sum); 77 int i=left,j=mid+1,k=left; 78 while(i<=mid&&j<=right) 79 arr[i]<=arr[j]? 80 tmp[k++]=arr[i++]: 81 tmp[k++]=arr[j++],sum+=mid-i+1;//Sum Is Used To Count The Reverse Alignment 82 while(i<=mid)tmp[k++]=arr[i++]; 83 while(j<=right)tmp[k++]=arr[j++]; 84 R(i,left,right)arr[i]=tmp[i]; 85 } 86 Iv Bucket_Sort(int a[],int left,int right){ 87 int mx=0; 88 R(i,left,right) 89 Cmax(mx,a[i]),++tot[a[i]]; 90 ++mx; 91 while(mx--) 92 while(tot[mx]--) 93 a[right--]=mx; 94 } 95 */ 96 int n,m,a[N],sum_start,tmp[N]; 97 Iv Merge_Sort(int arr[],int left,int right,int &sum){ 98 if(left>=right)return; 99 int mid=left+right>>1; 100 Merge_Sort(arr,left,mid,sum),Merge_Sort(arr,mid+1,right,sum); 101 int i=left,j=mid+1,k=left; 102 while(i<=mid&&j<=right) 103 arr[i]<=arr[j]? 104 tmp[k++]=arr[i++]: 105 (tmp[k++]=arr[j++],sum+=mid-i+1);//Sum Is Used To Count The Reverse Alignment 106 while(i<=mid)tmp[k++]=arr[i++]; 107 while(j<=right)tmp[k++]=arr[j++]; 108 R(i,left,right)arr[i]=tmp[i]; 109 } 110 int main(){ 111 int n; 112 while(scanf("%d %d",&n,&m)!=EOF,n,m){ 113 sum_start=0; 114 int sum_end,num_cnt=0; 115 R(i,1,n) 116 R(j,1,m){ 117 int num=read(); 118 !num? 119 sum_end=i: 120 a[++num_cnt]=num; 121 } 122 Merge_Sort(a,1,num_cnt,sum_start); 123 D_e(sum_start); 124 sum_end=n-sum_end; 125 if(m&1)sum_end=0; 126 (sum_start&1)==(sum_end&1)? 127 printf("YES "): 128 printf("NO "); 129 } 130 return 0; 131 } 132 /* 133 Note: 134 When Commas Are Used In Trinary Operators, Parentheses Shoule Be Used. 135 Error: 136 None. 137 */