Searching a particular word in a matrix of characters

问题

I was trying to search for a particular word in a matrix of characters through C but was unable to come to a fixed solution.

For ex: Suppose I have to search for the word INTELLIGENT in a matrix of characters (3*9) (Once you have picked a character from the matrix to form a sentence, you cannot pick it again to form the same sentence.There is a path from any cell to all its neighboring cells. A neighbor may share an edge or a corner.)

IIIINN.LI  
....TTEGL  
.....NELI

Output: YES (the word INTELLIGENT can be found) Can anybody please give a solution to the above problem !!!!

回答1:

#include <stdio.h>

char Matrix[3][9] = {
    { 'I','I','I','I','N','N','.','L','I'},
    { '.','.','.','.','T','T','E','G','L'},
    { '.','.','.','.',',','N','E','L','I'}
};
char Choice[3][9] = { { 0 }, { 0 }, { 0 } };
const char WORD[] = "INTELLIGENT";
const int  Len = sizeof(WORD)-1;
int Path[sizeof(WORD)-1] = { 0 };

char get(int row, int col){
    if(1 > col || col > 9) return '\0';
    if(1 > row || row > 3) return '\0';
    if(Choice[row-1][col-1] || Matrix[row-1][col-1] == '.')
        return '\0';
    else
        return Matrix[row-1][col-1];
}

#define toLoc(r, c) (r)*10+(c)
#define getRow(L) L/10
#define getCol(L) L%10

int search(int loc, int level){
    int r,c,x,y;
    char ch;
    if(level == Len) return 1;//find it
    r = getRow(loc);
    c = getCol(loc);
    ch = get(r,c);
    if(ch == 0 || ch != WORD[level]) return 0;
    Path[level]=toLoc(r,c);
    Choice[r-1][c-1] = 'v';//marking
    for(x=-1;x<=1;++x){
        for(y=-1;y<=1;++y){
            if(search(toLoc(r+y,c+x), level + 1)) return 1;
        }
    }
    Choice[r-1][c-1] = '\0';//reset
    return 0;
}

int main(void){
    int r,c,i;
    for(r=1;r<=3;++r){
        for(c=1;c<=9;++c){
            if(search(toLoc(r,c), 0)){
                printf("YES\nPath:");
                for(i=0;i<Len;++i){
                    printf("(%d,%d)", getRow(Path[i]), getCol(Path[i]));
                }
                printf("\n");
                return 0;
            }
        }
    }
    printf("NO\n");
    return 0;
}

回答2:

Use a depth first search.

You can do this using a recursive algorthm. Find all the (unused) places containing the first letter then see if it is possible to find the rest of the word on the remaining board by starting from one of the adjacent squares.

回答3:

I think this is what you mean..... Though it seems simpler to what you currently have been offered, so I may have misunderstood the question.

I use Numpy to reshape an arbitrary array into a single list of letters, then we create a mask of the search term and a copy of the input list. I tick off each letter to search for while updating the mask.


import numpy as np
import copy

def findInArray(I,Word):
    M=[list(x) for x in I]

    M=list(np.ravel(M))

    print "Letters to start: %s"%"".join(M)

    Mask=[False]*len(Word)

    T = copy.copy(M)

    for n,v in enumerate(Word):
        try:
            p=T.index(v)
        except ValueError:
            pass
        else:
            T[p]=''
            Mask[n]=True

    print "Letters left over: %s"%"".join(T)            
    if all(Mask):print "Found %s"%Word
    else:print "%s not Found"%Word

    print "\n"

    return all(Mask)


I=["IIIINN.LI","....TTEGL",".....NELI"]

findInArray(I,"INTEL")
findInArray(I,"INTELLIGENT")
findInArray(I,"INTELLIGENCE")

Example output

Letters to start: IIIINN.LI....TTEGL.....NELI
Letters left over: IIIN.I....TGL.....NELI
Found INTEL

Letters to start: IIIINN.LI....TTEGL.....NELI
Letters left over: II.I.........NLI
Found INTELLIGENT

Letters to start: IIIINN.LI....TTEGL.....NELI
Letters left over: II.I....T.....NLI
INTELLIGENCE not Found

回答4:

#include <stdio.h>

#define ROW 1
#define COL 11

char Matrix[ROW][COL] = { { 'I','N','T','E','L','L','I','G','E', 'N', 'T'} };
char Choice[ROW][COL] = { { 0 } };
const char WORD[] = "INTELLIGENT";
const int  Len = sizeof(WORD)-1;
int Path[sizeof(WORD)-1] = { 0 };

char get(int row, int col){
    if(1 > col || col > COL) return '\0';
    if(1 > row || row > ROW) return '\0';
    if(Choice[row-1][col-1] || Matrix[row-1][col-1] == '.')
        return '\0';
    else
        return Matrix[row-1][col-1];
}

#define toLoc(r, c) (r)*16+(c)
#define getRow(L) L/16
#define getCol(L) L%16

int search(int loc, int level){
    int r,c,x,y;
    char ch;
    if(level == Len) return 1;//find it
    r = getRow(loc);
    c = getCol(loc);
    ch = get(r,c);
    if(ch == 0 || ch != WORD[level]) return 0;
    Path[level]=toLoc(r,c);
    Choice[r-1][c-1] = 'v';//marking
    for(x=-1;x<=1;++x){
        for(y=-1;y<=1;++y){
            if(search(toLoc(r+y,c+x), level + 1)) return 1;
        }
    }
    Choice[r-1][c-1] = '\0';//reset
    return 0;
}

int main(void){
    int r,c,i;
    for(r=1;r<=ROW;++r){
        for(c=1;c<=COL;++c){
            if(search(toLoc(r,c), 0)){
                printf("YES\nPath:");
                for(i=0;i<Len;++i){
                    printf("(%d,%d)", getRow(Path[i]), getCol(Path[i]));
                }
                printf("\n");
                return 0;
            }
        }
    }
    printf("NO\n");
    return 0;
}

来源：https://stackoverflow.com/questions/10647950/searching-a-particular-word-in-a-matrix-of-characters

标签

matrix