题目链接：见这里
C. The Labyrinth
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

You are given a rectangular field of n × m cells. Each cell is either empty or impassable (contains an obstacle). Empty cells are marked with ‘.’, impassable cells are marked with ‘*’. Let’s call two empty cells adjacent if they share a side.

Let’s call a connected component any non-extendible set of cells such that any two of them are connected by the path of adjacent cells. It is a typical well-known definition of a connected component.

For each impassable cell (x, y) imagine that it is an empty cell (all other cells remain unchanged) and find the size (the number of cells) of the connected component which contains (x, y). You should do it for each impassable cell independently.

The answer should be printed as a matrix with n rows and m columns. The j-th symbol of the i-th row should be “.” if the cell is empty at the start. Otherwise the j-th symbol of the i-th row should contain the only digit —- the answer modulo 10. The matrix should be printed without any spaces.

To make your output faster it is recommended to build the output as an array of n strings having length m and print it as a sequence of lines. It will be much faster than writing character-by-character.

As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.
Input

The first line contains two integers n, m (1 ≤ n, m ≤ 1000) — the number of rows and columns in the field.

Each of the next n lines contains m symbols: “.” for empty cells, “*” for impassable cells.
Output

Print the answer as a matrix as described above. See the examples to precise the format of the output.
Examples
Input

3 3
.
.*.
.

Output

3.3
.5.
3.3

Input

4 5
*..
..*
...
..*

Output

46..3
..732
.6.4.
5.4.3

Note

In first example, if we imagine that the central cell is empty then it will be included to component of size 5 (cross). If any of the corner cell will be empty then it will be included to component of size 3 (corner).

题意：给你n行m列的矩阵，矩阵*表示障碍，.表示空地。对于每一个障碍，让你输出去掉这个障碍之后，这个点所在的连通块的大小是多少。答案需要%10。
解题方法：这个块的联通块大小，实际上是由四个部分所组成的，上下左右的连通块加在一起就好了，但是有一个问题，就是你有可能加重复，所以我们需要有个vis处理。之前比赛的时候用set写的，然后又改成带权dsu写了一发。

bfs + dsu:

//CF 616C
//GTMD dsu
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1150;
//const int maxm = 1e6+70;//WA
const int maxm = 1010050;//AC
char s[maxn][maxn];
int dir[4][2] = {{0, 1}, {0, -1}, {-1, 0}, {1, 0}};
int n, m, vis[maxm];
int getid(int x, int y){return x*m + y;}
namespace dsu{
    int fa[maxm], num[maxm];
    inline void init(){for(int i = 0; i <= getid(n, m) + 5; i++) fa[i] = i, num[i] = 1;}
    inline int find_set(int x){if(x == fa[x]) return x; else return fa[x] = find_set(fa[x]);}
    inline void union_set(int x, int y){int fx = find_set(x), fy = find_set(y); if(fx != fy){fa[fx] = fy; num[fy] = num[fx] + num[fy];}}
}
using namespace dsu;

int main()
{
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) scanf("%s", s[i] + 1);
    init();
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            if(s[i][j] == '*') continue;
            for(int k = 0; k < 4; k++){
                int tx = i + dir[k][0], ty = j + dir[k][1];
                if(tx < 1 || tx > n || ty < 1 || ty > m || s[tx][ty] == '*') continue;
                union_set(getid(tx, ty), getid(i, j));
            }
        }
    }
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            if(s[i][j] == '.') printf(".");
            else{
                int ans = 1;
                for(int k = 0; k < 4; k++){
                    int tx = i + dir[k][0], ty = j + dir[k][1];
                    if(tx < 1 || tx > n || ty < 1 || ty > m || s[tx][ty] == '*') continue;
                    int idx = find_set(getid(tx, ty));
                    if(vis[idx] == 0){
                        vis[idx] = 1;
                        ans += num[idx];
                    }
                }
                printf("%d", ans%10);
                for(int k = 0; k < 4; k++){
                    int tx = i + dir[k][0], ty = j + dir[k][1];
                    if(tx < 1 || tx > n || ty < 1 || ty > m || s[tx][ty] == '*') continue;
                    int idx = find_set(getid(tx, ty));
                    vis[idx] = 0;
                }
            }
        }
        puts("");
    }
}

bfs+set

const int maxn = 1005;
#define Read(x) scanf("%d",&x)
char G[maxn][maxn];
int cnt[maxn*maxn];
int vis[maxn][maxn];
int dir[4][2]={{0,1},{0,-1},{-1,0},{1,0}};
int n,m;
set <int> S;
bool check(int x,int y)
{
    if(!vis[x][y]&&G[x][y]!='*'&&x>=0&&x<n&&y>=0&&y<m)
        return true;
    return false;
}
void dfs(int x,int y,int id)//标记联通块
{
    vis[x][y]=id;
    cnt[id]++;
    for(int k=0; k<4; k++)
    {
        int dx = x+dir[k][0];
        int dy = y+dir[k][1];
        if(check(dx,dy))
        {
            dfs(dx,dy,id);
        }
    }
}
int main()
{
    int id;
    Read(n),Read(m);
    for(int i=0;i<n;i++)
    {
        scanf("%s",G[i]);
    }
    id=0;
    memset(cnt,0,sizeof(cnt));
    for(int i=0; i<n; i++)
    {
        for(int j=0; j<m; j++)
        {
            if(G[i][j]=='.'&&!vis[i][j])
            {
                dfs(i,j,++id);
            }
        }
    }
    for(int i=0 ;i<n; i++)
    {
        for(int j=0; j<m; j++)
        {
            if(G[i][j]=='*')
            {
                int ans=0;
                for(int k=0;k<4;k++)
                {
                    int dx = i+dir[k][0];
                    int dy = j+dir[k][1];
                    S.insert(vis[dx][dy]);
                }
                for(set<int>::iterator it = S.begin();it != S.end(); it++)
                {
                    ans+=cnt[*it];
                }
                S.clear();
                ans++;ans%=10;
                printf("%d",ans);
            }
            else
                printf(".");
        }
        puts("");
    }
    return 0;
}