Word Puzzles
Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 10782 Accepted: 4076 Special Judge Description
Word puzzles are usually simple and very entertaining for all ages. They are so entertaining that Pizza-Hut company started using table covers with word puzzles printed on them, possibly with the intent to minimise their client's perception of any possible delay in bringing them their order.
Even though word puzzles may be entertaining to solve by hand, they may become boring when they get very large. Computers do not yet get bored in solving tasks, therefore we thought you could devise a program to speedup (hopefully!) solution finding in such puzzles.
The following figure illustrates the PizzaHut puzzle. The names of the pizzas to be found in the puzzle are: MARGARITA, ALEMA, BARBECUE, TROPICAL, SUPREMA, LOUISIANA, CHEESEHAM, EUROPA, HAVAIANA, CAMPONESA.
Your task is to produce a program that given the word puzzle and words to be found in the puzzle, determines, for each word, the position of the first letter and its orientation in the puzzle.
You can assume that the left upper corner of the puzzle is the origin, (0,0). Furthemore, the orientation of the word is marked clockwise starting with letter A for north (note: there are 8 possible directions in total).
Input
The first line of input consists of three positive numbers, the number of lines, 0 < L <= 1000, the number of columns, 0 < C <= 1000, and the number of words to be found, 0 < W <= 1000. The following L input lines, each one of size C characters, contain the word puzzle. Then at last the W words are input one per line.
Output
Your program should output, for each word (using the same order as the words were input) a triplet defining the coordinates, line and column, where the first letter of the word appears, followed by a letter indicating the orientation of the word according to the rules define above. Each value in the triplet must be separated by one space only.
Sample Input
20 20 10 QWSPILAATIRAGRAMYKEI AGTRCLQAXLPOIJLFVBUQ TQTKAZXVMRWALEMAPKCW LIEACNKAZXKPOTPIZCEO FGKLSTCBTROPICALBLBC JEWHJEEWSMLPOEKORORA LUPQWRNJOAAGJKMUSJAE KRQEIOLOAOQPRTVILCBZ QOPUCAJSPPOUTMTSLPSF LPOUYTRFGMMLKIUISXSW WAHCPOIYTGAKLMNAHBVA EIAKHPLBGSMCLOGNGJML LDTIKENVCSWQAZUAOEAL HOPLPGEJKMNUTIIORMNC LOIUFTGSQACAXMOPBEIO QOASDHOPEPNBUYUYOBXB IONIAELOJHSWASMOUTRK HPOIYTJPLNAQWDRIBITG LPOINUYMRTEMPTMLMNBO PAFCOPLHAVAIANALBPFS MARGARITA ALEMA BARBECUE TROPICAL SUPREMA LOUISIANA CHEESEHAM EUROPA HAVAIANA CAMPONESASample Output
0 15 G 2 11 C 7 18 A 4 8 C 16 13 B 4 15 E 10 3 D 5 1 E 19 7 C 11 11 H
【分析】
听说TRIE+爆搜能过，于是我就爆搜了。
代码如下：
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring> 4 #include<iostream> 5 #include<algorithm> 6 #include<queue> 7 using namespace std; 8 #define Maxn 1010 9 #define Maxl 1010 10 11 char s[Maxl][Maxl]; 12 char ss[Maxl][Maxl]; 13 int dx[10]={0,1,1,0,-1,-1,-1,0,1}, 14 dy[10]={0,0,-1,-1,-1,0,1,1,1}; 15 int ans[Maxn][2],dr[Maxn]; 16 int n,m; 17 18 struct node 19 { 20 int son[30],cnt,fail; 21 int num,rt,p; 22 }t[Maxn*1000];int tot; 23 24 void upd(int x) 25 { 26 t[x].cnt=0;t[x].p=0; 27 memset(t[x].son,0,sizeof(t[x].son)); 28 } 29 30 void read_trie(int x) 31 { 32 scanf("%s",ss[x]+1); 33 int len=strlen(ss[x]+1); 34 int now=0; 35 for(int j=1;j<=len;j++) ss[0][j]=ss[x][len-j+1]; 36 for(int j=1;j<=len;j++) 37 { 38 int ind=ss[0][j]-'A'+1; 39 if(!t[now].son[ind]) 40 t[now].son[ind]=++tot,upd(tot); 41 now=t[now].son[ind]; 42 if(j==len) t[now].cnt++,t[now].p=x; 43 } 44 } 45 46 void dfs(int x,int y,int dir,int now) 47 { 48 if(t[now].p!=0) 49 { 50 int q=t[now].p; 51 ans[q][0]=x;ans[q][1]=y;dr[q]=dir; 52 t[now].p=0; 53 } 54 if(x+dx[dir]<1||x+dx[dir]>n||y+dy[dir]<1||y+dy[dir]>m) return; 55 int a=t[now].son[s[x+dx[dir]][y+dy[dir]]-'A'+1]; 56 if(a!=0) 57 dfs(x+dx[dir],y+dy[dir],dir,a); 58 } 59 60 void init() 61 { 62 int q; 63 scanf("%d%d%d",&n,&m,&q); 64 for(int i=1;i<=n;i++) 65 { 66 scanf("%s",s[i]+1); 67 } 68 tot=0;upd(0); 69 for(int i=1;i<=q;i++) 70 { 71 read_trie(i); 72 } 73 for(int i=0;i<=n+1;i++) 74 for(int j=0;j<=m+1;j++) 75 { 76 for(int k=1;k<=8;k++) 77 { 78 dfs(i,j,k,0); 79 } 80 } 81 for(int i=1;i<=q;i++) 82 { 83 printf("%d %d %c ",ans[i][0]-1,ans[i][1]-1,dr[i]+'A'-1); 84 } 85 } 86 87 int main() 88 { 89 init(); 90 return 0; 91 }
2016-06-19 14:09:06
后来还是打了一下AC自动机，嗯，感觉for那里还是超级慢，因为重叠那部分不好搞，就这样吧。
代码如下：
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring> 4 #include<iostream> 5 #include<algorithm> 6 #include<queue> 7 using namespace std; 8 #define Maxn 1010 9 #define Maxl 1010 10 11 char s[Maxl][Maxl]; 12 char ss[Maxl][Maxl]; 13 int dx[10]={0,1,1,0,-1,-1,-1,0,1}, 14 dy[10]={0,0,-1,-1,-1,0,1,1,1}; 15 int ans[Maxn][2],dr[Maxn]; 16 int n,m,qy; 17 18 struct node 19 { 20 int son[30],cnt,fail; 21 int num,rt; 22 }t[Maxn*1000];int tot; 23 24 int p[Maxn]; 25 26 void upd(int x) 27 { 28 t[x].cnt=0;/*t[x].p=0;*/ 29 memset(t[x].son,0,sizeof(t[x].son)); 30 } 31 32 void read_trie(int x) 33 { 34 scanf("%s",ss[x]+1); 35 int len=strlen(ss[x]+1); 36 int now=0; 37 for(int j=1;j<=len;j++) ss[0][j]=ss[x][len-j+1]; 38 for(int j=1;j<=len;j++) 39 { 40 int ind=ss[0][j]-'A'+1; 41 if(!t[now].son[ind]) 42 t[now].son[ind]=++tot,upd(tot); 43 now=t[now].son[ind]; 44 if(j==len) t[now].cnt++,p[x]=now; 45 } 46 } 47 48 queue<int > q; 49 void build_AC() 50 { 51 int i,j,x,y; 52 while(!q.empty()) q.pop(); 53 q.push(0); 54 while(!q.empty()) 55 { 56 x=q.front(); 57 y=t[x].fail; 58 for(j=1;j<=26;j++) 59 { 60 if(t[x].son[j]) 61 { 62 q.push(t[x].son[j]); 63 t[t[x].son[j]].fail=x?t[y].son[j]:x; 64 } 65 else t[x].son[j]=t[y].son[j]; 66 } 67 q.pop(); 68 } 69 } 70 71 void check(int x,int y,int dir,int now) 72 { 73 if(now==0) return; 74 for(int i=1;i<=qy;i++) if(p[i]==now) 75 { 76 ans[i][0]=x; 77 ans[i][1]=y; 78 dr[i]=dir; 79 } 80 now=t[now].fail; 81 check(x,y,dir,now); 82 } 83 84 void dfs(int x,int y,int dir,int now) 85 { 86 now=t[now].son[s[x][y]-'A'+1]; 87 check(x,y,dir,now); 88 x=x+dx[dir],y=y+dy[dir]; 89 if(x<1||x>n||y<1||y>m) return; 90 dfs(x,y,dir,now); 91 92 } 93 94 void init() 95 { 96 scanf("%d%d%d",&n,&m,&qy); 97 for(int i=1;i<=n;i++) 98 { 99 scanf("%s",s[i]+1); 100 } 101 tot=0;upd(0); 102 for(int i=1;i<=qy;i++) 103 { 104 read_trie(i); 105 } 106 build_AC(); 107 for(int i=1;i<=n;i++) 108 for(int k=1;k<=8;k++) 109 { 110 dfs(i,1,k,0); 111 dfs(i,m,k,0); 112 } 113 for(int j=1;j<=m;j++) 114 for(int k=1;k<=8;k++) 115 { 116 dfs(1,j,k,0); 117 dfs(n,j,k,0); 118 } 119 for(int i=1;i<=qy;i++) 120 { 121 printf("%d %d %c ",ans[i][0]-1,ans[i][1]-1,dr[i]+'A'-1); 122 } 123 } 124 125 int main() 126 { 127 init(); 128 return 0; 129 }
2016-06-21 17:14:17