poj2386 poj3620

2386.Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14109		Accepted: 7179

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.

Source

USACO 2004 November

#include <stdio.h>
#include<memory.h>
char a[111][111];
int m,n,f;
void dfs(int i,int j)
{
    if(a[i][j]=='.'||i<0||j<0||i>m-1||j>n-1) return ;
    a[i][j]='.';
    dfs(i,j+1);
    dfs(i,j-1);
    dfs(i+1,j);
    dfs(i+1,j-1);
    dfs(i+1,j+1);
    dfs(i-1,j);
    dfs(i-1,j+1);
    dfs(i-1,j-1);
}
int main()
{
    scanf("%d%d",&m,&n);
    f=0;
    memset(a,0,sizeof(a));
    for(int i=0;i<m;i++)
        scanf("%s",a[i]);
    for(int i=0;i<m;i++)
        for(int j=0;j<n;j++)
        {
            if(a[i][j]=='W')
            {
                f++;
                dfs(i,j);
            }
        }
    printf("%d\n",f);
    return 0;
}

3620.Avoid The Lakes

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5347		Accepted: 2865

Description

Farmer John's farm was flooded in the most recent storm, a fact only aggravated by the information that his cows are deathly afraid of water. His insurance agency will only repay him, however, an amount depending on the size of the largest "lake" on his farm.

The farm is represented as a rectangular grid with N (1 ≤ N ≤ 100) rows and M (1 ≤ M ≤ 100) columns. Each cell in the grid is either dry or submerged, and exactlyK (1 ≤ K ≤ N × M) of the cells are submerged. As one would expect, a lake has a central cell to which other cells connect by sharing a long edge (not a corner). Any cell that shares a long edge with the central cell or shares a long edge with any connected cell becomes a connected cell and is part of the lake.

Input

* Line 1: Three space-separated integers: N, M, and K
* Lines 2..K+1: Line i+1 describes one submerged location with two space separated integers that are its row and column: R and C

Output

* Line 1: The number of cells that the largest lake contains.　

Sample Input

3 4 5
3 2
2 2
3 1
2 3
1 1

Sample Output

4

Source

USACO 2007 November Bronze

判断上下左相通的cell的个数的max，DFS

#include <cstdio>
#include <iostream>
#include <memory.h>
using namespace std;
int a[110][110],vis[110][110];
int n,m,k;
int dfs(int i,int j)
{
    if(i<1||j<1||i>n||j>m||vis[i][j]!=0||a[i][j]==0)
    return 0;
    vis[i][j]=1;
    int t=1;
    t+=dfs(i-1,j);
    t+=dfs(i+1,j);
    t+=dfs(i,j+1);
    t+=dfs(i,j-1);
    return t;
}
int main()
{
    while(scanf("%d%d%d",&n,&m,&k)!=EOF)
    {
        int i,j,p,q,max;
        memset(a,0,sizeof(a));
        memset(vis,0,sizeof(vis));
        max=0;
        for(i=0;i<k;i++)
        {
            scanf("%d%d",&p,&q);
            a[p][q]=1;
        }
        for(i=1;i<=n;i++)
        for(j=1;j<=m;j++)
        if(a[i][j]&&!vis[i][j])
        {
            int sum=dfs(i,j);
            if(sum>max)
            max=sum;
        }
        printf("%d\n",max);
    }
    return 0;
}