Word Puzzles
| Time Limit: 5000MS |
|
Memory Limit: 65536K |
| Total Submissions: 6968 |
|
Accepted: 2651 |
|
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
CAMPONESA
Sample 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
Source
基本多串匹配
給定一個字母矩陣
在給定一些單詞，問這些單詞在矩陣中的位置和方向，
囧呆了，單詞構建trie樹，
把矩陣中的每一個8個方向之一的字符串看作文章
然后去做自動機
只需要枚舉周圍一圈的點的方向即可（想了好久才明白，自己去琢磨），
看來自己對ac自動機還不是理解的很透徹，得再看幾遍
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <cassert>
#include <iostream>
#include <sstream>
#include <fstream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <algorithm>
#include <iomanip>
#define maxn 1005
using namespace std;
int n,m,w;
char s1[maxn][maxn];
char str[maxn];
int dx[8][2]=
{{-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1}};
struct node
{
int next[26];
int count;
int fail;
void init()
{
memset(next,-1,sizeof(next));
count=0;fail=0;
}
}s[5000005];
struct result
{
int x,y,d;
}res[maxn];
int sind;
int q[500005],head,tail;
int len[maxn];
void cas_init()
{
s[0].init();
sind=1;
}
void ins(char srt[],int k)
{
int len=strlen(str);
int i,j,ind;
ind=0;
for(i=0;i<len;i++)
{
j=str[i]-'A';
if(s[ind].next[j]==-1)
{
s[sind].init();
s[ind].next[j]=sind++;
}
ind=s[ind].next[j];
}
s[ind].count=k;
}
void make_fail()
{
head=0;tail=0;
int i,ind,ind_f;
for(i=0;i<26;i++)
{
if(s[0].next[i]!=-1)
{
q[++tail]=s[0].next[i];
}
}
while(head<tail)
{
ind=q[++head];
for(i=0;i<26;i++)
{
if(s[ind].next[i]!=-1)
{
q[++tail]=s[ind].next[i];
ind_f=s[ind].fail;
while(ind_f>0&&s[ind_f].next[i]==-1)
ind_f=s[ind_f].fail;
if(s[ind_f].next[i]!=-1)
ind_f=s[ind_f].next[i];
s[s[ind].next[i]].fail=ind_f;
}
}
}
}
bool check(int x,int y)
{
return (x>=0&&x<n&&y>=0&&y<m);
}
void fd(int x1,int y1,int d)
{
int di,i,ind,p,x,y;
ind=0;x=x1;y=y1;
for(;check(x,y);x+=dx[d][0],y+=dx[d][1])
{
i=s1[x][y]-'A';
while(ind>0&&s[ind].next[i]==-1)
ind =s[ind].fail;
if(s[ind].next[i]!=-1)
{
ind=s[ind].next[i];
p=ind;
//printf("%d\n",ind);
while(p>0&&s[p].count!=-1)
{
int tmp=s[p].count;
res[tmp].x=x-(len[tmp]-1)*dx[d][0];
res[tmp].y=y-(len[tmp]-1)*dx[d][1];
res[tmp].d=d;
//cout<<s[p].count<<" "<<x1<<" "<<y1<<" "d<<endl;
s[p].count=-1;
p=s[p].fail;
}
}
}
}
int main()
{
scanf("%d%d%d",&n,&m,&w);
for(int i=0;i<n;i++) scanf("%s",s1[i]);
cas_init();
for(int i=1;i<=w;i++)
{
scanf("%s",str);
len[i]=strlen(str);
ins(str,i);
}
make_fail();
for(int i=0;i<m;i++)
{
fd(0,i,3);fd(0,i,4);fd(0,i,5);
fd(n-1,i,7);fd(n-1,i,0);fd(n-1,i,1);
}
for(int i=0;i<n;i++)
{
fd(i,0,1);fd(i,0,2);fd(i,0,3);
fd(i,m-1,5);fd(i,m-1,6);fd(i,m-1,7);
}
for(int i=1;i<=w;i++)
printf("%d %d %c\n",res[i].x,res[i].y,res[i].d+'A');
return 0;
}
這個題說的還是比較模糊的了，沒有說明多個匹配的情況怎么辦，
但是它special judge了，應該是輸出任意一中情況即可