Codeforces Round #470 (rated, Div. 2, based on VK Cup 2018 Round 1)A. Protect Sheep

http://codeforces.com/contest/948/problem/A

 
A. Protect Sheep

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
Copy
6 6
..S...
..S.W.
.S....
..W...
...W..
......
output
Yes
..SD..
..SDW.
.SD...
.DW...
DD.W..
......
input
Copy
1 2
SW
output
No
input
Copy
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.

 

题意就是给一个N*C矩阵,让你往空白的'.'的地方放狗,用来把羊'S'和狼'W'分隔开来,狗的数量不限,输出任意解。所以只要遍历整个矩阵,如果有S和W相邻则不可行,反之可行,所有空白处都放上狗。

复制代码
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
#define eps 0.00000001
#define pn printf("\n")
using namespace std;
typedef long long ll;

int n,m;
char s[550][550];
int to[4][2] = {0,1,0,-1,1,0,-1,0};

int main()
{
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++)
        scanf("%s",s[i]);
    int fl = 0;
    for(int i=0;i<n;i++)
        for(int j=0;j<m;j++)
        {
            if(s[i][j] == 'S')
            {
                for(int k=0;k<4;k++)
                {
                    int tx = i + to[k][0], ty = j + to[k][1];
                    if(tx >=0 && tx < n && ty >= 0 && ty < m)
                    {
                        if(s[tx][ty] == 'W')
                        {
                            fl = 1; break;
                        }
                    }
                }
            }
        }
    if(fl) printf("No\n");
    else
    {
        printf("Yes\n");
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(s[i][j] == '.') printf("D");
                else printf("%c",s[i][j]);
            }
            printf("\n");
        }
    }
}
复制代码

 

 

posted @   HazelNuto  阅读(434)  评论(0编辑  收藏  举报
编辑推荐:
· DeepSeek 解答了困扰我五年的技术问题
· 为什么说在企业级应用开发中,后端往往是效率杀手?
· 用 C# 插值字符串处理器写一个 sscanf
· Java 中堆内存和栈内存上的数据分布和特点
· 开发中对象命名的一点思考
阅读排行:
· DeepSeek 解答了困扰我五年的技术问题。时代确实变了!
· PPT革命!DeepSeek+Kimi=N小时工作5分钟完成?
· What?废柴, 还在本地部署DeepSeek吗?Are you kidding?
· DeepSeek企业级部署实战指南:从服务器选型到Dify私有化落地
· 程序员转型AI:行业分析
点击右上角即可分享
微信分享提示