中石油12203-Darker and Darker

题目描述

You are given a grid of squares with H horizontal rows and W vertical columns, where each square is painted white or black. HW characters from A11 to AHW represent the colors of the squares. Aij is # if the square at the i-th row from the top and the j-th column from the left is black, and Aij is . if that square is white.
We will repeatedly perform the following operation until all the squares are black:
Every white square that shares a side with a black square, becomes black.
Find the number of operations that will be performed. The initial grid has at least one black square.
Constraints
·1≤H,W≤1000
·Aij is # or ..
·The given grid has at least one black square.

输入

Input is given from Standard Input in the following format:
H W
A11A12...A1W
:
AH1AH2...AHW

输出

Print the number of operations that will be performed.

样例输入

复制样例数据

3 3
...
.#.
...

样例输出

提示

After one operation, all but the corners of the grid will be black. After one more operation, all the squares will be black.

就不翻译了

思路：广度优先搜索模板题

自己写的代码：

 1 #include<bits/stdc++.h>2 using namespace std;3 typedef long long ll;4 int vis[1005][1005];5 char arr[1005][1005];6 int main(){7    int n,m;cin>>n>>m;8    for(int i=1;i<=n;i++){9     scanf("%s",arr[i]+1);
10    }
11    memset(vis,-1,sizeof(vis));
12    typedef pair<int,int>P;
13    queue<P>que;
14    for(int i=1;i<=n;i++){
15     for(int j=1;j<=m;j++){
16         if(arr[i][j]=='#') {
17                 vis[i][j]=0;
18                 que.push(P(i,j));
19         }
20     }
21    }
22    int maxx=0;
23    while(!que.empty()){
24     P p=que.front();que.pop();
25     if(p.first-1>0&&vis[p.first-1][p.second]==-1){
26         vis[p.first-1][p.second]=vis[p.first][p.second]+1;
27         que.push(P(p.first-1,p.second));
28         maxx=max(maxx,vis[p.first][p.second]+1);
29     }
30     if(p.first+1<=n&&vis[p.first+1][p.second]==-1){
31         vis[p.first+1][p.second]=vis[p.first][p.second]+1;
32         que.push(P(p.first+1,p.second));
33         maxx=max(maxx,vis[p.first][p.second]+1);
34     }
35     if(p.second-1>0&&vis[p.first][p.second-1]==-1){
36            vis[p.first][p.second-1]=vis[p.first][p.second]+1;
37         que.push(P(p.first,p.second-1));
38         maxx=max(maxx,vis[p.first][p.second]+1);
39     }
40     if(p.second+1<=m&&vis[p.first][p.second+1]==-1){
41            vis[p.first][p.second+1]=vis[p.first][p.second]+1;
42         que.push(P(p.first,p.second+1));
43         maxx=max(maxx,vis[p.first][p.second]+1);
44     }
45    }
46    cout<<maxx<<endl;
47 return 0;
48 }

大神写的代码：

 1 #include<bits/stdc++.h>2 using namespace std;3 const int maxn=1e5+5;4 const int dx[]={1,-1,0,0};5 const int dy[]={0,0,1,-1};6 7 int n,m,ans;8 char s[1111][1111];9 struct node { int x,y,step; };
10 queue<node> que;
11 void bfs() {
12     while (!que.empty()) {
13         node now=que.front(); que.pop();
14         ans=max(ans,now.step);
15         for (int i=0;i<4;i++) {
16             node nex=node{now.x+dx[i],now.y+dy[i],now.step+1};
17             if (s[ nex.x ][ nex.y ]=='.') {
18                 s[ nex.x ][ nex.y ]='#';
19                 que.push(nex);
20             }
21         }
22     }
23 }
24 int main () {
25     scanf("%d%d",&n,&m);
26     for (int i=1;i<=n;i++)
27         for (int j=1;j<=m;j++) {
28             scanf(" %c",&s[i][j]);
29             if (s[i][j]=='#')
30                 que.push(node{i,j,0});
31         }
32     bfs();
33     printf("%d\n",ans);
34     return 0;
35 }

总结：刚开始的时候竟然往dfs上面想了，看来是好久没见bfs的题了，都不会做了。另外代码跟神犇写的差距太大了，要好好学习别人的代码呀。

　　　　　蒟蒻一枚；