China mining freshman race H Lu Shen See the Island "Bfs/dfs the link block/connected block area at the boundary, this connecting block is invalid"

Time limit: C/s 1 sec, other languages 2 seconds
Space limitations: C/C + + 32768K, other languages 65536K
64bit IO Format:%lld topic description N*m now has a map of the size of the map, marked by the land (denoted by "#") and the Sea (with "." The number of islands on this map is now being calculated by the goddess. It is known that the island is made up of connected blocks of land, that is, the upper, lower, left, right, upper left, upper right, lower left, and lower right side of a land, which constitute connected blocks, etc. In addition, the island's detailed definition is as follows: 1. The islands must be surrounded by oceans. 2, if the connected block has any area at the map boundary, then the connected block is not an island. Input Description:

Line 1th enters two integer n,m, which represents the length and width of the map.
Line 2-n+1, enter m characters per line, character "#" for Land, "." Represents the ocean.
Data guarantee: 0<n,m≤200

Output Description:

Outputs a single line of integers representing the number of islands.

Example 1 input

3 3....# ....

Output

1

Description

Only the median of 1 land is an island, so the number of islands = 1

Example 2 input

3 3#...# ....

Output

0

Description

The middle connecting block has an area at the boundary, so not the island, number of islands = 0.

"Analysis": the BFS solves the connected block, noting that in the solution process, if the connected block region is at the boundary, it is not valid to record this connecting block, but the BFS operation still needs to be completed.
"Code":

#include <bits/stdc++.h>using namespacestd;Const intMAXN = -;Charmp[ -][ -];intvis[maxn][maxn]={0};intdir[][2] ={{1,0},{0,1},{-1,0},{0,-1},{-1,-1},{-1,1},{1,-1},{1,1}};intn,m,flag=0;voidDfsintXinty) {Vis[x][y]=1; if(x==0|| y==0|| x==n-1|| y==m-1) Flag=1;  for(intI=0;i<8; i++)    {        inttx=x+dir[i][0]; intty=y+dir[i][1]; if(mp[tx][ty]=='#'&&tx>=0&&ty>=0) {Mp[tx][ty]='.';        DFS (Tx,ty); }    }}intMain () {intsum=0; CIN>>n>>m;  for(intI=0; i<n;i++) Cin>>Mp[i];  for(intI=0; i<n;i++)    {         for(intj=0; j<m;j++)        {            if(mp[i][j]=='#') {flag=0;                DFS (I,J); if(flag==0) Sum++; } }} cout<<sum<<Endl; return 0;}

DFS

#include <bits/stdc++.h>using namespacestd;Const intmaxn= About; CharMP[MAXN][MAXN];intVIS[MAXN][MAXN];Const intdir[8][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1}}; intn,m,ans=0;voidBFsintXinty) {    if(mp[x][y]=='.') {Vis[x][y]=2; return; }    if(x==0|| y==0|| x==n-1|| y==m-1) {Vis[x][y]=2; //return;} queue<pair<int,int> >Q;    Q.push (Make_pair (x, y));  while(!Q.empty ()) {        intnowx=Q.front (). First; intnowy=Q.front (). Second; if(nowx==0|| nowy==0|| nowx==n-1|| nowy==m-1) {Vis[x][y]=2; //return;} q.pop ();  for(intI=0;i<8;++i) {            intx1=nowx+dir[i][0]; inty1=nowy+dir[i][1]; if(vis[x1][y1]==0&&mp[x1][y1]=='#') {Vis[x1][y1]=1;            Q.push (Make_pair (x1,y1)); }        }    }} intMain () { while(cin>>n>>L) {memset (MP,0,sizeof(MP)); memset (Vis,false,sizeof(VIS)); Ans=0;  for(intI=0; i<n;++i) Cin>>Mp[i];  for(intI=0; i<n;++i) {             for(intj=0; j<m;++j) {                if(!Vis[i][j]) {Vis[i][j]=1;                    BFS (I,J); if(vis[i][j]==1) {ans++; } }}} cout<<ans<<Endl; }     return 0;}

BFS

China mining freshman race H Lu Shen See the Island "Bfs/dfs the link block/connected block area at the boundary, this connecting block is invalid"