DFS:POJ3620-Avoid The Lakes(求最基本的联通块)

Avoid The Lakes

Time Limit: 1000MS

Memory Limit: 65536K

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 exactly K (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

1. Line 1: Three space-separated integers: N, M, and K
2. 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

1. 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

---

好久没写dfs了有点手生。

---

#include<stdio.h>
#include<cstring>
const int maxn = 110;
char maps[maxn][maxn];
int dir[4][2] = {1,0,-1,0,0,1,0,-1};
bool vis[maxn][maxn];
int n,m,k,ans,num;

void pre_maps()//自己建立一个地图
{
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            maps[i][j] = '*';
}

bool check(int x,int y)//检查是否跑到了地图外面
{
    if(x<1 || y<1 || x>n || y>m)
        return false;
    return true;
}

void dfs(int x,int y)
{
    vis[x][y] = true;
    num++;//找到一个算一个
    if(num > ans)
        ans = num;
    for(int i=0;i<4;i++)
        if(check(x+dir[i][0],y+dir[i][1]) && maps[x+dir[i][0]][y+dir[i][1]] == '#' && !vis[x+dir[i][0]][y+dir[i][1]])
            dfs(x+dir[i][0],y+dir[i][1]);//四个方向直接找，注意标记
    return;
}

void DFS()
{
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            if(maps[i][j] == '#' && !vis[i][j])
            {
                num = 0;
                dfs(i,j);
            }
}

int main()
{
    while(scanf("%d%d%d",&n,&m,&k) != EOF)
    {
        ans = 0;
        memset(vis,0,sizeof(vis));
        pre_maps();
        while(k--)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            maps[a][b] = '#';
        }

        DFS();
        printf("%d\n",ans);
    }
    return 0;
}