题目链接：https://codeforces.com/contest/948/problem/A
Bob is a farmer. He has a large pasture with many sheep. Recently, he has lost some of them due to wolf attacks. He thus decided to place some shepherd dogs in such a way that all his sheep are protected.
The pasture is a rectangle consisting of R × C cells. Each cell is either empty, contains a sheep, a wolf or a dog. Sheep and dogs always stay in place, but wolves can roam freely around the pasture, by repeatedly moving to the left, right, up or down to a neighboring cell. When a wolf enters a cell with a sheep, it consumes it. However, no wolf can enter a cell with a dog.
Initially there are no dogs. Place dogs onto the pasture in such a way that no wolf can reach any sheep, or determine that it is impossible. Note that since you have many dogs, you do not need to minimize their number.
Input
First line contains two integers R (1 ≤ R ≤ 500) and C (1 ≤ C ≤ 500), denoting the number of rows and the numbers of columns respectively.
Each of the following R lines is a string consisting of exactly C characters, representing one row of the pasture. Here, ‘S’ means a sheep, ‘W’ a wolf and ‘.’ an empty cell.
Output
If it is impossible to protect all sheep, output a single line with the word “No”.
Otherwise, output a line with the word “Yes”. Then print R lines, representing the pasture after placing dogs. Again, ‘S’ means a sheep, ‘W’ a wolf, ‘D’ is a dog and ‘.’ an empty space. You are not allowed to move, remove or add a sheep or a wolf.
If there are multiple solutions, you may print any of them. You don’t have to minimize the number of dogs.
Examples
Input
6 6
…S…
…S.W.
.S…
…W…
…W…
…
Output
Yes
…SD…
…SDW.
.SD…
.DW…
DD.W…
…
Input
1 2
SW
Output
No
Input
5 5
.S…
…S.
S…
…S.
.S…
Output
Yes
.S…
…S.
S.D…
…S.
.S…
Note
In the first example, we can split the pasture into two halves, one containing wolves and one containing sheep. Note that the sheep at (2,1) is safe, as wolves cannot move diagonally.
In the second example, there are no empty spots to put dogs that would guard the lone sheep.
In the third example, there are no wolves, so the task is very easy. We put a dog in the center to observe the peacefulness of the meadow, but the solution would be correct even without him.

题意：一个牧场是由R×C细胞组成的矩形。每个细胞都是空的，包含一只绵羊，一只狼或一只狗。绵羊和狗总是留在原地，但是狼可以在牧场周围自由地漫游，通过反复向左，向右，向上或向下移动到相邻的细胞。要你放狗保护棉羊，狗的数量可以不用考虑，如果放狗能保护每一个绵羊则输出Yes，否则，绵羊身边必有狼，输出No。
解题思路：我们只要考虑每个绵羊上 下 左 右是否存在狼，是则必是No，否则打印Yes，再把是"."的位置变成"D"再输出。

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
char m[510][510];
int r,c;
int main()
{
    scanf("%d%d",&r,&c);
    for(int i=1;i<=r;i++)
      for(int j=1;j<=c;j++)
      cin>>m[i][j];//输入

    for(int i=1;i<=r;i++)
      for(int j=1;j<=c;j++)
      {
          if(m[i][j]=='S')
          {
              if(m[i-1][j]=='W'||m[i][j-1]=='W'||m[i+1][j]=='W'||m[i][j+1]=='W')//判断一下羊身边是否存在狼
            {
                printf("No");
                return 0;//两个for循环我不能break；那么我直接结束程序，就是这么任性；
            }
        }
      }
      printf("Yes\n");
    for(int i=1;i<=r;i++){
      for(int j=1;j<=c;j++)
      {
          if(m[i][j]=='.') printf("D");//如果是空地，则全变成狗。
        else printf("%c",m[i][j]);
      }
      cout<<endl;
    }
      return 0;
}