Codeforces 980B（思维）

题面：

The city of Fishtopia can be imagined as a grid of $4$ rows and an odd number of columns. It has two main villages; the first is located at the top-left cell $(1, 1)$ , people who stay there love fishing at the Tuna pond at the bottom-right cell $(4, n)$ . The second village is located at $(4, 1)$ and its people love the Salmon pond at $(1, n)$ .

The mayor of Fishtopia wants to place $k$ hotels in the city, each one occupying one cell. To allow people to enter the city from anywhere, hotels should not be placed on the border cells.

A person can move from one cell to another if those cells are not occupied by hotels and share a side.

Can you help the mayor place the hotels in a way such that there are equal number of shortest paths from each village to its preferred pond?

Input

The first line of input contain two integers, $n$ and $k$ ( $3 \leq n \leq 99$ , $0 \leq k \leq 2 \times (n - 2)$ ), $n$ is odd, the width of the city, and the number of hotels to be placed, respectively.

Output

Print "YES", if it is possible to place all the hotels in a way that satisfies the problem statement, otherwise print "NO".

If it is possible, print an extra $4$ lines that describe the city, each line should have $n$ characters, each of which is "#" if that cell has a hotel on it, or "." if not.

    Examples 
  

      input 
     
       Copy 
     
7 2

      output 
     
       Copy 
     
YES
.......
.#.....
.#.....
.......

      input 
     
       Copy 
     
5 3

      output 
     
       Copy 
     
YES
.....
.###.
.....
.....

题面描述:

给你一个4*n的矩阵，A村庄在左上角的点（1，1），它要到达右下角C点（4,n)，B村庄在左下角（4，1），它要达到右上角D（1，n）。先我们要建k个旅馆，要求所建的旅店的位置不能改变A到C的最短路径的数量以及B到D的最短路径。要求你输出满足条件的矩阵。

题目分析：

因为A到C和B到D的最短距离必定为他们之间的曼哈顿距离。而要满足所有最短路的数量都相同，则酒店所设立的位置必定是要将整个图关于x轴或者y轴对称。

对于这个题目，因为题目要求，我们不能在边界上建立酒店，因此我们只能在大小为2*(n-2)的矩阵中建立酒店。而题目所给的k数据范围是[0,(n-2)*2]的，因此不难发现对于这道题来说，只要合理安排酒店的位置，必定是可以将k个酒店都安排好的。

于是，我们进一步观察分析，可以知道，对于这个2*（n-2）的矩阵，要使得图关于x或者y轴对称，则当k为偶数的时候，只需要不断填充第2行第3行即可；而当k为奇数的时候，我们只需要在这个2*（n-2）的矩阵的中心开始加点，并从两边往中间加酒店即可。

代码：

#include <bits/stdc++.h>
using namespace std;
string str[5];
int main()
{
    int n,k;
    cin>>n>>k;
    for(int i=0;i<4;i++){
        for(int j=0;j<n;j++){
            str[i]+='.';
        }
    }
    if(k%2==0){
        k/=2;
        for(int i=1;i<=2;i++){
            for(int j=1;j<=k;j++){
                str[i][j]='#';
            }
        }
    }
    else{
        str[1][n/2]='#';
        if(k!=1){
            int tmp=n/2;
            if(k<=n-2){
                for(int i=1;i<=(k-1)/2;i++){
                    str[1][i]='#';
                    str[1][n-1-i]='#';
                }
            }
            else{
                for(int i=1;i<=n-2;i++) str[1][i]='#';
                k-=n-2;
                for(int i=1;i<=k/2;i++){
                    str[2][i]='#';
                    str[2][n-1-i]='#';
                }
            }

        }
    }
    puts("YES");
    for(int i=0;i<4;i++){
        cout<<str[i]<<endl;
    }
}