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

2

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：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int vis[1005][1005];
char arr[1005][1005];
int main(){
   int n,m;cin>>n>>m;
   for(int i=1;i<=n;i++){
    scanf("%s",arr[i]+1);
   }
   memset(vis,-1,sizeof(vis));
   typedef pair<int,int>P;
   queue<P>que;
   for(int i=1;i<=n;i++){
    for(int j=1;j<=m;j++){
        if(arr[i][j]=='#') {
                vis[i][j]=0;
                que.push(P(i,j));
        }
    }
   }
   int maxx=0;
   while(!que.empty()){
    P p=que.front();que.pop();
    if(p.first-1>0&&vis[p.first-1][p.second]==-1){
        vis[p.first-1][p.second]=vis[p.first][p.second]+1;
        que.push(P(p.first-1,p.second));
        maxx=max(maxx,vis[p.first][p.second]+1);
    }
    if(p.first+1<=n&&vis[p.first+1][p.second]==-1){
        vis[p.first+1][p.second]=vis[p.first][p.second]+1;
        que.push(P(p.first+1,p.second));
        maxx=max(maxx,vis[p.first][p.second]+1);
    }
    if(p.second-1>0&&vis[p.first][p.second-1]==-1){
           vis[p.first][p.second-1]=vis[p.first][p.second]+1;
        que.push(P(p.first,p.second-1));
        maxx=max(maxx,vis[p.first][p.second]+1);
    }
    if(p.second+1<=m&&vis[p.first][p.second+1]==-1){
           vis[p.first][p.second+1]=vis[p.first][p.second]+1;
        que.push(P(p.first,p.second+1));
        maxx=max(maxx,vis[p.first][p.second]+1);
    }
   }
   cout<<maxx<<endl;
return 0;
}

大神写的代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+5;
const int dx[]={1,-1,0,0};
const int dy[]={0,0,1,-1};

int n,m,ans;
char s[1111][1111];
struct node { int x,y,step; };
queue<node> que;
void bfs() {
    while (!que.empty()) {
        node now=que.front(); que.pop();
        ans=max(ans,now.step);
        for (int i=0;i<4;i++) {
            node nex=node{now.x+dx[i],now.y+dy[i],now.step+1};
            if (s[ nex.x ][ nex.y ]=='.') {
                s[ nex.x ][ nex.y ]='#';
                que.push(nex);
            }
        }
    }
}
int main () {
    scanf("%d%d",&n,&m);
    for (int i=1;i<=n;i++)
        for (int j=1;j<=m;j++) {
            scanf(" %c",&s[i][j]);
            if (s[i][j]=='#')
                que.push(node{i,j,0});
        }
    bfs();
    printf("%d\n",ans);
    return 0;
}

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；