題干：

求每個*能夠到達的格子數量，只有.可以走（四個方向擴展），結果mod 10，替換 * 后輸出。

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).

解題報告：

? ?直接對每個'.'進行搜索看連通塊，然后對每一個'*'直接求答案就行了，注意不要算重復，所以需要set去判重。（因為有可能四個方向屬于同一個聯通塊，所以不判重就會加四次）

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair using namespace std; const int MAX = 1000 + 5; char maze[MAX][MAX]; int ans[MAX][MAX]; bool vis[MAX][MAX]; int f[MAX*MAX]; int num[MAX*MAX]; int n,m; int nx[4] = {0,1,0,-1}; int ny[4] = {1,0,-1,0}; int getf(int v) {return v == f[v] ? v : f[v] = getf(f[v]); } void merge(int u,int v) {int t1 = getf(u);int t2 = getf(v);f[t2] = t1; } int get(int x,int y) {return (x-1)*m +y; } void go(int x,int y) {for(int k = 0; k<4; k++) {int tx = x + nx[k];int ty = y + ny[k];if(tx < 1 || tx > n || ty < 1 || ty > m) continue;if(maze[tx][ty] == '*' || vis[tx][ty]) continue;merge(get(x,y),get(tx,ty));vis[tx][ty] = 1;go(tx,ty);} } int main() {cin>>n>>m;for(int i = 0; i<=n*m; i++) {f[i] = i;}for(int i = 1; i<=n; i++) {scanf("%s",maze[i]+1);}for(int i = 1; i<=n; i++) {for(int j = 1; j<=m; j++) {if(maze[i][j] == '.' && vis[i][j] == 0) {vis[i][j] = 1;//別忘這步！！雖然這題不會WA但是不寫是錯的。go(i,j);} }}for(int i = 1; i<=n; i++) {for(int j = 1; j<=m; j++) {if(maze[i][j] == '*') continue;num[getf(get(i,j))]++;}}for(int i = 1; i<=n; i++) {for(int j = 1; j<=m; j++) {if(maze[i][j] == '.') ans[i][j] = '.';else {int cnt = 0;set<int> ss;for(int k = 0; k<4; k++) {int tx = i + nx[k];int ty = j + ny[k];if(maze[tx][ty] == '*')continue;if(tx < 1 || tx > n || ty < 1 || ty > m) continue;ss.insert(getf(get(tx,ty)));}auto it = ss.begin();for(;it!=ss.end();++it) cnt += num[*it];ans[i][j] = cnt;}}}for(int i = 1; i<=n; i++) {for(int j = 1; j<=m; j++) {if(maze[i][j] == '.') printf(".");else printf("%d",(ans[i][j]+1)%10);}puts("");}return 0 ; }

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