POJ2386 Lake Counting(dfs)
2017-01-11 16:21
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Lake Counting
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
Sample Output
3
dfs代码如下:
#include<cstdio>
using namespace std;
const int MAX_N=100;
const int MAX_M=100;
int N,M;
char field[MAX_N][MAX_M+1];
void dfs(int x,int y){
field[x][y]='.';
for(int dx=-1;dx<=1;dx++){
for(int dy=-1;dy<=1;dy++){
int nx=x+dx,ny=y+dy;
if(0<=nx&&nx<N&&0<=ny&&ny<M&&field[nx][ny]=='W') dfs(nx,ny);
}
}
return ;
}
void solve(){
int res =0;
for(int i=0;i<N;i++){
for(int j=0;j<M;j++){
if(field[i][j]=='W'){
dfs(i,j);
res++;
}
}
}
printf("%d\n",res);
}
int main(){
while(~scanf("%d%d",&N,&M)){
for(int i=0;i<N;i++){
scanf("%s",field[i]);
}
solve();
}
return 0;
}
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 17917 | Accepted: 9069 |
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
dfs代码如下:
#include<cstdio>
using namespace std;
const int MAX_N=100;
const int MAX_M=100;
int N,M;
char field[MAX_N][MAX_M+1];
void dfs(int x,int y){
field[x][y]='.';
for(int dx=-1;dx<=1;dx++){
for(int dy=-1;dy<=1;dy++){
int nx=x+dx,ny=y+dy;
if(0<=nx&&nx<N&&0<=ny&&ny<M&&field[nx][ny]=='W') dfs(nx,ny);
}
}
return ;
}
void solve(){
int res =0;
for(int i=0;i<N;i++){
for(int j=0;j<M;j++){
if(field[i][j]=='W'){
dfs(i,j);
res++;
}
}
}
printf("%d\n",res);
}
int main(){
while(~scanf("%d%d",&N,&M)){
for(int i=0;i<N;i++){
scanf("%s",field[i]);
}
solve();
}
return 0;
}
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