POJ3620Avoid The Lakes

题目描述

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.

与UVA572不同的是，本题不需求连通块的数量，需要的是最大的连通分量的大小，而且连通的规则也改为了只能上下左右才可以。
日常安利本题我的博客

思路分析

这个题的思路是这样的，除了之前说过的求连通分量的数组外，额外加了一个存连通分量大小的数组，最后将这个数组有大到小排序，输出数组的第一个元素即可。
这个题代码部分DFS函数没写好，因为其他的错误以为DFS边界写错了，于是添加了冗余代码。
DFS判断边界情况只需要考虑当前情况即可。引以为戒。

代码

#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;

#define mem(a) memset(a,0,sizeof(a))
#define ll long long

const double eps = 1e-8;
const int maxn = 110;//须填写
const int inf = 0x3f3f3f3f;

int n,m;
int pic[maxn][maxn];
int vis[maxn][maxn];
int num[maxn*maxn];

bool cmp(int a,int b)
{
      return a>b;

}

void dfs(int x,int y,int s)
{
    if(x<1||y<1||x>n||y>m)
        return ;
    if(!vis[x-1][y]&&pic[x-1][y]==1&&x-1>=1)
    {
        vis[x-1][y]=s;
        num[s]++;
        dfs(x-1,y,s);
    }
    if(!vis[x+1][y]&&pic[x+1][y]==1&&x+1<=n)
    {
        vis[x+1][y]=s;
        num[s]++;
        dfs(x+1,y,s);
    }
    if(!vis[x][y-1]&&pic[x][y-1]==1&&y-1>=1)
    {
        vis[x][y-1]=s;
        num[s]++;
        dfs(x,y-1,s);
    }
    if(!vis[x][y+1]&&pic[x][y+1]==1&&y+1<=m)
    {
        vis[x][y+1]=s;
        num[s]++;
        dfs(x,y+1,s);
    }
}

int main()
{
    //freopen("in.txt", "r", stdin);
    //freopen("sample.out", "w", stdout);
    mem(vis);
    mem(pic);
    mem(num);
    int k,x,y;
    int results = 0;
    scanf("%d%d%d",&n,&m,&k);
    for(int i=0;i