hihoCoder题库1094Lost in the City

题目

时间限制: 10000ms

单点时限: 1000ms

内存限制: 256MB

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.

输入

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.

输出

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.

样例输入

8 8
...HSH..
...HSM..
...HST..
...HSPP.
PPGHSPPT
PPSSSSSS
..MMSHHH
..MMSH..
SSS
SHG
SH.

样例输出

5 4

我的代码（Wrong Answer）

#include
#include
#include
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]){           //右转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];      //复制数组
		}
	}
	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

代码一开始就是wrong answer,然后我就去查到了这位博主的代码

http://blog.csdn.net/u014355480/article/details/43459097，

然后改进了我的旋转和输入部分的代码，仔细对比之后也觉得没有差，但是他的能AC我的就是WA，搞不太懂，先放在这里。

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问题已经解决了，是我没有考虑边界

最后一部分的代码贴在下面（结果已A）

for(i=1;i

map[i][j]的起始搜索只能从(1,1)开始，到（N-1，M-1）处，边界没有意义。