Where's Waldorf?

Where's Waldorf?

时间: 1ms        内存:64M

描述:

Given a m by n grid of letters, ($1 \leq m,n \leq 20$), and a list of words, find the location in the grid at which the word can be found.

A word matches a straight, uninterrupted line of letters in the grid.A word can match the letters in the grid regardless of case (i.e. upper and lower case letters are to be treated as the same). The matching can be done in any of the eight directions either horizontally, vertically or diagonally through the grid.

输入:

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

The input begins with a pair of integers, m followed by n, $1 \leq m,n \leq 50$in decimal notation on a single line. The next m lines contain n letters each; this is the grid of letters in which the words of the list must be found. The letters in the grid may be in upper or lower case. Following the grid of letters, another integer

k appears on a line by itself ($1 \leq k \leq 20$). The next k
lines of input contain the list of words to search for, one word perline. These words may contain upper and lower case letters only (no spaces, hyphens or other non-alphabetic characters).

输出:

For each word in each test case, output a pair of integers representing its location in the corresponding grid. The integers must be separated by a single space. The first integer is the line in the grid where the first letter of the given word can be found (1 represents the topmost line in the grid, and m represents the bottommost line). The second integer is the column in the grid where the first letter of the given word can be found (1 represents the leftmost column in the grid, and n represents the rightmost column in the grid). If a word can be found more than once in the grid, then output the location of the uppermost occurrence of the word (i.e., the occurrence which places the first letter of the word closest to the top of the grid). If two or more words are uppermost, output the leftmost of these occurrences. All words can be found at least once in the grid.

The output of two consecutive cases must be separated by a blank line.

示例输入:

1

8 11
abcDEFGhigg
hEbkWalDork
FtyAwaldORm
FtsimrLqsrc
byoArBeDeyv
Klcbqwikomk
strEBGadhrb
yUiqlxcnBjf
4
Waldorf
Bambi
Betty
Dagbert

示例输出:

2 5
2 3
1 2
7 8

提示:

参考答案(内存最优[1824]):

#include<stdio.h>
#include<string.h>
char str[1000][1000];
char key[100][1000];
int main () {
    int n,i,j,k,u,h,w,num,len,ok,index;
    while (scanf("%d",&n) != EOF) {
        for (u=0;u<n;u++) {
            if (u) {
                printf("\n");
            }
            scanf("%d%d",&h,&w);
            for (i=0;i<h;i++) {
                scanf("%s",str[i]);
                len = strlen(str[i]);
                for (j=0;j<len;j++){
                    if (str[i][j]>='A'&&str[i][j]<='Z') {
                        str[i][j] +='a'-'A';
                    }
                }
            }
            scanf("%d",&num);
            for(i=0;i<num;i++) {
                scanf("%s",key[i]);
                len = strlen(key[i]);
                for (j=0;j<len;j++){
                    if (key[i][j]>='A'&&key[i][j]<='Z') {
                        key[i][j] +='a'-'A';
                    }
                }
            }
            len = strlen(key[i = 0]);
            for (j=1;j<=h;j++) {
                for (k=1;k<=w;k++) {
                    ok = 1;
                    if (str[j-1][k-1]==key[i][0]&&k>=len) {
                        for (index=0;index<len;index++) {
                            if (str[j-1][k-index-1]!=key[i][index]) ok = 0;
                        }
                        if (ok) {
                            printf("%d %d\n",j,k);
                            j = 0;
                            i++;
                            if (i<num)
                            len = strlen(key[i]);
                            break;
                        }
                        if (j>=len) {
                            ok = 1;
                            for (index=0;index<len;index++) {
                                if (str[j-index-1][k-index-1]!=key[i][index]) ok = 0;
                            }
                            if (ok) {
                                printf("%d %d\n",j,k);
                                j = 0;
                                i++;
                                if (i<num)
                                len = strlen(key[i]);
                                break;
                            }
                        } else if (h-j+1>=len) {
                            ok = 1;
                            for (index=0;index<len;index++) {
                                if (str[j+index-1][k-index-1]!=key[i][index]) ok = 0;
                            }
                            if (ok) {
                                printf("%d %d\n",j,k);
                                j = 0;
                                i++;
                                if (i<num)
                                len = strlen(key[i]);
                                break;
                            }
                        }
                    } if (str[j-1][k-1]==key[i][0]&&w-k+1>=len) {
                        ok = 1;
                        for (index=0;index<len;index++) {
                            if (str[j-1][k+index-1]!=key[i][index]) ok = 0;
                        }
                        if (ok) {
                            printf("%d %d\n",j,k);
                            j = 0;
                            i++;
                            if (i<num)
                            len = strlen(key[i]);
                            break;
                        }
                        if (j>=len) {
                            ok = 1;
                            for (index=0;index<len;index++) {
                                if (str[j-index-1][k+index-1]!=key[i][index]) ok = 0;
                                }
                                if (ok) {
                                    printf("%d %d\n",j,k);
                                    j = 0;
                                    i++;
                                    if (i<num)
                                    len = strlen(key[i]);
                                    break;
                                }
                        }
                        if (h-j+1>=len) {
                            ok = 1;
                            for (index=0;index<len;index++) {
                                if (str[j+index-1][k+index-1]!=key[i][index]) ok = 0;
                            }
                            if (ok) {
                                printf("%d %d\n",j,k);
                                j = 0;
                                i++;
                                if (i<num)
                                len = strlen(key[i]);
                                break;
                            }
                        }
                    } if (str[j-1][k-1]==key[i][0]&&j>=len) {
                        ok = 1;
                        for (index=0;index<len;index++) {
                            if (str[j-index-1][k-1]!=key[i][index]) ok = 0;
                        }
                        if (ok) {
                            printf("%d %d\n",j,k);
                            j = 0;
                            i++;
                            if (i<num)
                            len = strlen(key[i]);
                            break;
                        }
                    } if (str[j-1][k-1]==key[i][0]&&h-j+1>=len) {
                        ok = 1;
                        for (index=0;index<len;index++) {
                            if (str[j+index-1][k-1]!=key[i][index]) ok = 0;
                        }
                        if (ok) {
                            printf("%d %d\n",j,k);
                            j = 0;
                            i++;
                            if (i<num)
                            len = strlen(key[i]);
                            break;
                        }
                    }
                }
            }
        }
    }
    return 0;
}

参考答案(时间最优[4]):

#include<stdio.h>
#include<string.h>
char str[1000][1000];
char key[100][1000];
int main () {
    int n,i,j,k,u,h,w,num,len,ok,index;
    while (scanf("%d",&n) != EOF) {
        for (u=0;u<n;u++) {
            if (u) {
                printf("\n");
            }
            scanf("%d%d",&h,&w);
            for (i=0;i<h;i++) {
                scanf("%s",str[i]);
                len = strlen(str[i]);
                for (j=0;j<len;j++){
                    if (str[i][j]>='A'&&str[i][j]<='Z') {
                        str[i][j] +='a'-'A';
                    }
                }
            }
            scanf("%d",&num);
            for(i=0;i<num;i++) {
                scanf("%s",key[i]);
                len = strlen(key[i]);
                for (j=0;j<len;j++){
                    if (key[i][j]>='A'&&key[i][j]<='Z') {
                        key[i][j] +='a'-'A';
                    }
                }
            }
            len = strlen(key[i = 0]);
            for (j=1;j<=h;j++) {
                for (k=1;k<=w;k++) {
                    ok = 1;
                    if (str[j-1][k-1]==key[i][0]&&k>=len) {
                        for (index=0;index<len;index++) {
                            if (str[j-1][k-index-1]!=key[i][index]) ok = 0;
                        }
                        if (ok) {
                            printf("%d %d\n",j,k);
                            j = 0;
                            i++;
                            if (i<num)
                            len = strlen(key[i]);
                            break;
                        }
                        if (j>=len) {
                            ok = 1;
                            for (index=0;index<len;index++) {
                                if (str[j-index-1][k-index-1]!=key[i][index]) ok = 0;
                            }
                            if (ok) {
                                printf("%d %d\n",j,k);
                                j = 0;
                                i++;
                                if (i<num)
                                len = strlen(key[i]);
                                break;
                            }
                        } else if (h-j+1>=len) {
                            ok = 1;
                            for (index=0;index<len;index++) {
                                if (str[j+index-1][k-index-1]!=key[i][index]) ok = 0;
                            }
                            if (ok) {
                                printf("%d %d\n",j,k);
                                j = 0;
                                i++;
                                if (i<num)
                                len = strlen(key[i]);
                                break;
                            }
                        }
                    } if (str[j-1][k-1]==key[i][0]&&w-k+1>=len) {
                        ok = 1;
                        for (index=0;index<len;index++) {
                            if (str[j-1][k+index-1]!=key[i][index]) ok = 0;
                        }
                        if (ok) {
                            printf("%d %d\n",j,k);
                            j = 0;
                            i++;
                            if (i<num)
                            len = strlen(key[i]);
                            break;
                        }
                        if (j>=len) {
                            ok = 1;
                            for (index=0;index<len;index++) {
                                if (str[j-index-1][k+index-1]!=key[i][index]) ok = 0;
                                }
                                if (ok) {
                                    printf("%d %d\n",j,k);
                                    j = 0;
                                    i++;
                                    if (i<num)
                                    len = strlen(key[i]);
                                    break;
                                }
                        }
                        if (h-j+1>=len) {
                            ok = 1;
                            for (index=0;index<len;index++) {
                                if (str[j+index-1][k+index-1]!=key[i][index]) ok = 0;
                            }
                            if (ok) {
                                printf("%d %d\n",j,k);
                                j = 0;
                                i++;
                                if (i<num)
                                len = strlen(key[i]);
                                break;
                            }
                        }
                    } if (str[j-1][k-1]==key[i][0]&&j>=len) {
                        ok = 1;
                        for (index=0;index<len;index++) {
                            if (str[j-index-1][k-1]!=key[i][index]) ok = 0;
                        }
                        if (ok) {
                            printf("%d %d\n",j,k);
                            j = 0;
                            i++;
                            if (i<num)
                            len = strlen(key[i]);
                            break;
                        }
                    } if (str[j-1][k-1]==key[i][0]&&h-j+1>=len) {
                        ok = 1;
                        for (index=0;index<len;index++) {
                            if (str[j+index-1][k-1]!=key[i][index]) ok = 0;
                        }
                        if (ok) {
                            printf("%d %d\n",j,k);
                            j = 0;
                            i++;
                            if (i<num)
                            len = strlen(key[i]);
                            break;
                        }
                    }
                }
            }
        }
    }
    return 0;
}

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