题干：
Hongcow likes solving puzzles.
One day, Hongcow finds two identical puzzle pieces, with the instructions "make a rectangle" next to them. The pieces can be described by an n by m grid of characters, where the character 'X' denotes a part of the puzzle and '.' denotes an empty part of the grid. It is guaranteed that the puzzle pieces are one 4-connected piece. See the input format and samples for the exact details on how a jigsaw piece will be specified.
The puzzle pieces are very heavy, so Hongcow cannot rotate or flip the puzzle pieces. However, he is allowed to move them in any directions. The puzzle pieces also cannot overlap.
You are given as input the description of one of the pieces. Determine if it is possible to make a rectangle from two identical copies of the given input. The rectangle should be solid, i.e. there should be no empty holes inside it or on its border. Keep in mind that Hongcow is not allowed to flip or rotate pieces and they cannot overlap, i.e. no two 'X' from different pieces can share the same position.
Input
The first line of input will contain two integers n and m (1 ≤ n, m ≤ 500), the dimensions of the puzzle piece.
The next n lines will describe the jigsaw piece. Each line will have length m and will consist of characters '.' and 'X' only. 'X' corresponds to a part of the puzzle piece, '.' is an empty space.
It is guaranteed there is at least one 'X' character in the input and that the 'X' characters form a 4-connected region.
Output
Output "YES" if it is possible for Hongcow to make a rectangle. Output "NO" otherwise.
Examples
Input
2 3
XXX
XXX
Output
YES
Input
2 2
.X
XX
Output
NO
Input
5 5
.....
..X..
.....
.....
.....
Output
YES
Note
For the first sample, one example of a rectangle we can form is as follows
111222
111222
For the second sample, it is impossible to put two of those pieces without rotating or flipping to form a rectangle.
In the third sample, we can shift the first tile by one to the right, and then compose the following rectangle:
.....
..XX.
.....
.....
.....题目大意：
判断两块相同的有X的区域是否能拼成一个矩形。
解题报告：
首先分析出，只有矩形才能拼成矩形，因此判断给定的这个图形是否是一个矩形即可
（ps：这是暑假集训刚刚开始的组队题目，当时这个题看起来还无从下手，比赛时ac这个题的人也不是很多，但是现在看来就是个规规矩矩的水题小case了，，，看来水平真的在长进呀，想起两个月前迷茫的自己，我很开心。）
（ps：其实这应该不算个典型的字符串问题，，好吧都挂上了那就贴上这个标签吧）
AC代码：
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
char maze[501][501];
char ss[501];//模式空串
char ac[501];//模式匹配串
int main()
{
int n,m;
int num=0;
scanf("%d %d", &n, &m);
for(int i = 0; i<m; i++) {
ss[i]='.';
}
for(int i = 0; i<n; i++) {
scanf("%s",&maze[i]);
}
int flag=0;
for(int i = 0; i<n; i++) {
if(flag==0) {
if(!strcmp(ss,maze[i]) ) continue;
flag=1;//代表这个串不是空串了
for(int j = 0; j<m; j++) {
//代表这个串不是空串了
ac[j]=maze[i][j];
if(maze[i][j]=='X') num++;
}
}
else if(flag==1) {
if(!strcmp(maze[i],ss) ) {
printf("YES\n");
return 0 ;
}
if(!strcmp(maze[i],ac) ) {
continue;
}
if(strcmp(maze[i],ac) ) {
printf("NO\n");
return 0 ;
}
}
}
printf("YES\n");
return 0 ;
}