BFS
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <queue>
struct A
{
int x,y,time;
}t,now;
int map1[350][350];
int m,n;
int vis[350][350];
int dir[4][2]={-1,0,1,0,0,1,0,-1};
using namespace std;
int bfs(int x,int y)
{
memset(vis,0,sizeof(vis));
queue <A> q;
vis[x][y]=1;
now.x=x;
now.y=y;
now.time=0;
q.push(now);
while(!q.empty())
{
now=q.front();
q.pop();
if(map1[now.x][now.y]=='E')
{
return now.time;
}
for(int i=0;i<4;i++)
{
t.x=now.x+dir[i][0];
t.y=now.y+dir[i][1];
t.time=now.time+1;
if(t.x>=0&&t.x<n&&t.y>=0&&t.y<m&&vis[t.x][t.y]==0&&map1[t.x][t.y]!='#')
{
q.push(t);
vis[t.x][t.y]=1;
}
}
}
return -1;
}
int main()
{
int i,j,c=0,d=0;
scanf("%d%d",&n,&m);
for(i=0;i<n;i++)
{
getchar();
for(j=0;j<m;j++)
{
scanf(" %c",&map1[i][j]);
if(map1[i][j]=='S')
{
c=i;
d=j;
}
}
}
int ans=bfs(c,d);
printf("%d\n",ans);
return 0;
}
样例：
4 4
. . . .
. S . #
. # . .
. . . .DFS
G - Lake Counting
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status Practice POJ 2386
use MathJax to parse formulas
Description
Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.
Given a diagram of Farmer John's field, determine how many ponds he has.
Input
* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.
Output
* Line 1: The number of ponds in Farmer John's field.
Sample Input
10 12 W........WW. .WWW.....WWW ....WW...WW. .........WW. .........W.. ..W......W.. .W.W.....WW. W.W.W.....W. .W.W......W. ..W.......W.
Sample Output
3
Hint
OUTPUT DETAILS:
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
#include <iostream>
using namespace std;
char mapp[101][101];
int dir[8][2]={0,1,0,-1,-1,0,1,0,-1,1,1,1,-1,-1,1,-1};
int n,m;
void DFS(int x,int y)
{
if(x>=0&&x<m&&y>=0&&y<n&&mapp[x][y]=='W')
{
mapp[x][y]='.';
for(int i=0;i<8;i++)
DFS(x+dir[i][0],y+dir[i][1]);
}
return ;
}
int main()
{
int i,j,k;
cin>>m>>n;
for(i=0;i<m;i++)
for(j=0;j<n;j++)
cin>>mapp[i][j];
int ans=0;
for(i=0;i<m;i++)
for(j=0;j<n;j++)
if(mapp[i][j]=='W')
{
DFS(i,j);
ans++;
}
cout<<ans<<endl;
return 0;
}
