【37.50%】【codeforces 745B】Hongcow Solves A Puzzle

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.
…..
…..
…..

【题目链接】:http://codeforces.com/contest/745/problem/B

【题解】

题意有点迷;
大概是说那个所给的输入组成的一个X联通块只能整体移动吧.
然后问你两个这样的联通块能不能组成一个长方形。
因为什么都变不了。
所以只要在输入的那个图里面判断X是不是长方形就好了(中间不能有空的);
用bfs搞搞.

【完整代码】

#include <iostream>
#include <queue>
#include <cstring>
#include <cstdio>

using namespace std;

const int MAXN = 550;
const int dx[5] = {0,0,0,1,-1};
const int dy[5] = {0,1,-1,0,0};

int n,m,fz = 0,nq = 0;
int a[MAXN][MAXN];
queue <pair<int,int> > dl;

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    memset(a,0,sizeof(a));
    cin >> n >> m;
    for (int i = 1;i <= n;i++)
    {
        char key;
        for (int j = 1;j <=m;j++)
        {
            cin >> key;
            if (key == 'X')
                a[i][j] = 1;
            else
                a[i][j] = 0;
        }
    }
    for (int i = 1;i <= n;i++)
        for (int j = 1;j <= m;j++)
            if (a[i][j]==1)
            {
                int minx=i,maxx=i,miny=j,maxy=j;
                a[i][j] = 0;
                int now = 1;
                dl.push(make_pair(i,j));
                while (!dl.empty())
                {
                    int x = dl.front().first,y = dl.front().second;
                    dl.pop();
                    for (int p = 1;p <= 4;p++)
                    {
                        int tx = x+dx[p],ty = y+dy[p];
                        if (a[tx][ty])
                        {
                            now++;
                            minx = min(minx,tx);maxx = max(maxx,tx);
                            miny = min(miny,ty);maxy = max(maxy,ty);
                            a[tx][ty] = 0;
                            dl.push(make_pair(tx,ty));
                        }
                    }
                }
                int chang = maxy-miny+1;
                int kuan = maxx-minx+1;
                int total = chang*kuan;
                if (total == now)
                {
                    puts("YES");
                    return 0;
                }
                else
                {
                    puts("NO");
                    return 0;
                }
            }
    return 0;
}