題目
描述
Little Hi gets lost in the city. He does not know where he is. He does not know which direction is north.
Fortunately, Little Hi has a map of the city. The map can be considered as a grid of N*M blocks. Each block is numbered by a pair of integers. The block at the north-west corner is (1, 1) and the one at the south-east corner is (N, M). Each block is represented by a character, describing the construction on that block: '.' for empty area, 'P' for parks, 'H' for houses, 'S' for streets, 'M' for malls, 'G' for government buildings, 'T' for trees and etc.
Given the blocks of 3*3 area that surrounding Little Hi(Little Hi is at the middle block of the 3*3 area), please find out the position of him. Note that Little Hi is disoriented, the upper side of the surrounding area may be actually north side, south side, east side or west side.
輸入
Line 1: two integers, N and M(3 <= N, M <= 200).
Line 2~N+1: each line contains M characters, describing the city's map. The characters can only be 'A'-'Z' or '.'.
Line N+2~N+4: each line 3 characters, describing the area surrounding Little Hi.
輸出
Line 1~K: each line contains 2 integers X and Y, indicating that block (X, Y) may be Little Hi's position. If there are multiple possible blocks, output them from north to south, west to east.
8 8 ...HSH.. ...HSM.. ...HST.. ...HSPP. PPGHSPPT PPSSSSSS ..MMSHHH ..MMSH.. SSS SHG SH.樣例輸出
5 4我的代碼（Wrong Answer）
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
bool check(char a[][205],int i,int j,char b[][5]){
if(a[i-1][j-1]==b[0][0]){
if(a[i-1][j]==b[0][1]){
if(a[i-1][j+1]==b[0][2]){
if(a[i][j-1]==b[1][0]){
if(a[i][j+1]==b[1][2]){
if(a[i+1][j-1]==b[2][0]){
if(a[i+1][j]==b[2][1]){
if(a[i+1][j+1]==b[2][2]){
return true;
}
}
}
}
}
}
}
}
return false;
}
void change(char a[][5]){ //右轉90度
int i,j;
char b[3][3];
for(i=0;i<3;i++){
for(j=0;j<3;j++){
b[i][j]=a[i][j]; //複製數組
}
}
for(i=0;i<3;i++){
a[i][2]=b[0][i];
}
for(i=0;i<3;i++){
a[i][1]=b[1][i];
}
for(i=0;i<3;i++){
a[i][0]=b[2][i];
}
}
int main(){
int N,M,i,j;
char port[5][5],port1[5][5],port2[5][5],port3[5][5];
cin>>N>>M;
char map[205][205];
for(i=0;i<N;i++){
scanf("%s", map[i]);
}
for(i=0;i<3;i++){
scanf("%s", port[i]);
}
for(i=0;i<3;i++){ //port向右轉90度
for(j=0;j<3;j++){
port1[i][j]=port[i][j];
}
}
change(port1);
for(i=0;i<3;i++){ //port向右轉180度
for(j=0;j<3;j++){
port2[i][j]=port1[i][j];
}
}
change(port2);
for(i=0;i<3;i++){ //port向右轉270度
for(j=0;j<3;j++){
port3[i][j]=port2[i][j];
}
}
change(port3);
for(i=0;i<N;i++){
for(j=0;j<M;j++){
if(map[i][j]==port[1][1]){
if(check(map,i,j,port)||check(map,i,j,port1)||check(map,i,j,port2)||check(map,i,j,port3)){
cout<<i+1<<" "<<j+1<<endl;
}
}
}
}
//system("pause");
return 0;
}代碼一開始就是wrong answer,然後我就去查到了這位博主的代碼
http://blog.csdn.net/u014355480/article/details/43459097，
然後改進了我的旋轉和輸入部分的代碼，仔細對比之後也覺得沒有差，但是他的能AC我的就是WA，搞不太懂，先放在這裏。
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問題已經解決了，是我沒有考慮邊界
最後一部分的代碼貼在下面（結果已A）
for(i=1;i<N;i++){
for(j=1;j<M;j++){
if(map[i][j]==port[1][1]){
if(check(map,i,j,port)||check(map,i,j,port1)||check(map,i,j,port2)||check(map,i,j,port3)){
cout<<i+1<<" "<<j+1<<endl;
}
}
}
}