匹配行UVA 题目10010 - Where's Waldorf?
发一下牢骚和主题无关：
| Where's Waldorf? |
Given a
m
by
n
grid of letters, (
), 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.
Input
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, 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 ( ). The next k lines of input contain the list of words to search for, one word per line. These words may contain upper and lower case letters only (no spaces, hyphens or other non-alphabetic characters).
Output
For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.
For each word in the word list, a pair of integers representing the location of the corresponding word in the grid must be output. 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 the location which is output should correspond to the uppermost occurence of the word (i.e. the occurence which places the first letter of the word closest to the top of the grid). If two or more words are uppermost, the output should correspond to the leftmost of these occurences. All words can be found at least once in the grid.
Sample Input
1 8 11 abcDEFGhigg hEbkWalDork FtyAwaldORm FtsimrLqsrc byoArBeDeyv Klcbqwikomk strEBGadhrb yUiqlxcnBjf 4 Waldorf Bambi Betty Dagbert
Sample Output
2 5 2 3 1 2 7 8
Miguel Revilla
2000-08-22
【粗心】：
输入：
给你一个由字母成组的网格，M行N列。找寻一个单词在网格中的置位。一个单词匹配网格中联系不间断的字母。可以沿意任方向匹配，一共可以匹配八个方向。略忽大小写。
须要匹配的字符串有K个。
出输：
每组出输之间都一行空行。
m n：m代表匹配的最上面的行
n代表匹配的最下面的行
如果结果有多个，只出输匹配串。要求：匹配串的第一个字母必须是最高最左的。结果至少有一个。
【代码】：
/*********************************
* 期日：2013-4-23
* 作者：SJF0115
* 题号: 标题10010 - Where's Waldorf?
* 起源：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=12&page=show_problem&problem=951
* 结果：AC
* 起源：UVA
* 结总：
**********************************/
#include<stdio.h>
#include<string.h>
char Matrix[51][51];
char str[21],temp[21];
int StartR,StartC;
//M行 N列
int Match(int M,int N,int &StartR,int &StartC){
int i,j,k,flag;
StartR = 51,StartC = 51;
int len = strlen(str);
for(i = 0;i < M;i++){
for(j = 0;j < N;j++){
flag = 1;
//left - right
if(j + len <= N){
flag = 0;
for(k = 0;k < len;k++){
if(str[k] != Matrix[i][j+k]){
flag = 1;
break;
}
}
if(flag == 0){
if(StartR > i+1){
StartR = i+1;
StartC = j+1;
}
else if(StartR == i+1 && StartC > j+1){
StartR = i+1;
StartC = j+1;
}
}
}
//right - left
if(j - len + 1>= 0){
flag = 0;
for(k = 0;k < len;k++){
if(str[k] != Matrix[i][j-k]){
flag = 1;
break;
}
}
if(flag == 0){
if(StartR > i+1){
StartR = i+1;
StartC = j+1;
}
else if(StartR == i+1 && StartC > j+1){
StartR = i+1;
StartC = j+1;
}
}
}
//up - down
if(i + len <= M){
flag = 0;
for(k = 0;k < len;k++){
if(str[k] != Matrix[i+k][j]){
flag = 1;
break;
}
}
if(flag == 0){
if(StartR > i+1){
StartR = i+1;
StartC = j+1;
}
else if(StartR == i+1 && StartC > j+1){
StartR = i+1;
StartC = j+1;
}
}
}
//down - up
if(i - len + 1 >= 0){
flag = 0;
for(k = 0;k < len;k++){
if(str[k] != Matrix[i-k][j]){
flag = 1;
break;
}
}
if(flag == 0){
if(StartR > i+1){
StartR = i+1;
StartC = j+1;
}
else if(StartR == i+1 && StartC > j+1){
StartR = i+1;
StartC = j+1;
}
}
}
//right - up
if(j + len <= N && i - len + 1 >= 0){
flag = 0;
for(k = 0;k < len;k++){
if(str[k] != Matrix[i-k][j+k]){
flag = 1;
break;
}
}
if(flag == 0){
if(StartR > i+1){
StartR = i+1;
StartC = j+1;
}
else if(StartR == i+1 && StartC > j+1){
StartR = i+1;
StartC = j+1;
}
}
}
//right - down
if(j + len <= N && i + len <= M){
flag = 0;
for(k = 0;k < len;k++){
if(str[k] != Matrix[i+k][j+k]){
flag = 1;
break;
}
}
if(flag == 0){
if(StartR > i+1){
StartR = i+1;
StartC = j+1;
}
else if(StartR == i+1 && StartC > j+1){
StartR = i+1;
StartC = j+1;
}
}
}
//left - up
if(j - len + 1 >= 0 && i - len + 1 >= 0){
flag = 0;
for(k = 0;k < len;k++){
if(str[k] != Matrix[i-k][j-k]){
flag = 1;
break;
}
}
if(flag == 0){
if(StartR > i+1){
StartR = i+1;
StartC = j+1;
}
else if(StartR == i+1 && StartC > j+1){
StartR = i+1;
StartC = j+1;
}
}
}
//left - down
if(j - len + 1 >= 0 && i + len <= M){
flag = 0;
for(k = 0;k < len;k++){
if(str[k] != Matrix[i-k][j+k]){
flag = 1;
break;
}
}
if(flag == 0){
if(StartR > i+1){
StartR = i+1;
StartC = j+1;
}
else if(StartR == i+1 && StartC > j+1){
StartR = i+1;
StartC = j+1;
}
}
}
}//for j
}//for i
return 0;
}
int main (){
int i,j,Case,k,M,N;
//freopen("C:\\Users\\XIAOSI\\Desktop\\acm.txt","r",stdin);
while(scanf("%d",&Case) != EOF){
while(Case--){
scanf("%d %d",&M,&N);
//输入字符阵矩
for(i = 0;i < M;i++){
scanf("%s",temp);
for(j = 0;j < N;j++){
Matrix[i][j] = temp[j];
//转换为小写
if(Matrix[i][j] >= 'A' && Matrix[i][j] <= 'Z'){
Matrix[i][j] = Matrix[i][j] - 'A' + 'a';
}
}
}
scanf("%d",&k);
//待匹配串
for(i = 0;i < k;i++){
scanf("%s",str);
int len = strlen(str);
//转换为小写
for(j = 0;j < len;j++){
if(str[j] >= 'A' && str[j] <= 'Z'){
str[j] = str[j] - 'A' + 'a';
}
}
//printf("%s",str);
Match(M,N,StartR,StartC);
printf("%d %d\n",StartR,StartC);
}
//每组测试之间有空行
if(Case){
printf("\n");
}
}
}
return 0;
}
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