[USACO10OCT] Lake Counting S
题面翻译
由于近期的降雨,雨水汇集在农民约翰的田地不同的地方。我们用一个
N
×
M
(
1
≤
N
≤
100
,
1
≤
M
≤
100
)
N\times M(1\leq N\leq 100, 1\leq M\leq 100)
N×M(1≤N≤100,1≤M≤100) 的网格图表示。每个网格中有水(W
) 或是旱地(.
)。一个网格与其周围的八个网格相连,而一组相连的网格视为一个水坑。约翰想弄清楚他的田地已经形成了多少水坑。给出约翰田地的示意图,确定当中有多少水坑。
输入第 1 1 1 行:两个空格隔开的整数: N N N 和 M M M。
第
2
2
2 行到第
N
+
1
N+1
N+1 行:每行
M
M
M 个字符,每个字符是 W
或 .
,它们表示网格图中的一排。字符之间没有空格。
输出一行,表示水坑的数量。
题目描述
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.
输入格式
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.
输出格式
Line 1: The number of ponds in Farmer John’s field.
样例 #1
样例输入 #1
10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.
样例输出 #1
3
提示
OUTPUT DETAILS: There are three ponds: one in the upper left, one in the lower left, and one along the right side.
先贴一个bfs的模板
#include<iostream>
#include<cstdio>
using namespace std;
const int N=110;
const int flag[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
char c[N][N];
int a[N][N],queue[N*N][2];
int n,m,ans;
bool p[N][N];
void bfs(int x,int y)
{
int front=0,rear=2;
queue[1][0]=x,queue[1][1]=y;
while(front<rear-1)
{
++front;
x=queue[front][0];
y=queue[front][1];
for(int i=0;i<4;++i)
{
int x1=x+flag[i][0];
int y1=y+flag[i][1];
if(x1<1||x1>n||y1<1||y1>n||!a[x1][y1]||p[x1][y1]) continue;
p[x1][y1]=true;
queue[rear][0]=x1;
queue[rear++][1]=y1;
}
}
}
int main()
{
cin>>n>>m;
for(int i=1;i<=n;++i)
for(int j=1;j<=m;++j)
{
scanf(" %c",&c[i][j]);
if(c[i][j]=='W') a[i][j]=1;
}
for(int i=1;i<=n;++i)
for(int j=1;j<=m;++j)
if(a[i][j] && !p[i][j])
{
++ans;
bfs(i,j);
}
cout<<ans<<endl;
return 0;
}
再贴一个dfs的
#include<cstdio>
#include<cstring>
const int maxn=105;
char pic[maxn][maxn];
int m,n,idx[maxn][maxn];
void dfs(int r,int c,int id)
{
if(r<0||r>=m||c<0||c>=m) return;
if(idx[r][c]>0 || pic[r][c]!='W') return;
idx[r][c]=id;
for(int dr=-1;dr<=1;++dr)
for(int dc=-1;dc<=1;++dc)
if(dr!=0||dc!=0) dfs(r+dr,c+dc,id);
}
int main()
{
while(scanf("%d%d",&m,&n)==2 && m && n)
{
for(int i=0;i<m;++i) scanf("%s",pic[i]);
memset(idx,0,sizeof(idx));
int cnt=0;
for(int i=0;i<m;++i)
for(int j=0;j<n;++j)
if(idx[i][j]==0 && pic[i][j]=='W') dfs(i,j,++cnt);
printf("%d\n",cnt);
}
return 0;
}