POJ-3620 Avoid The Lakes（深搜）

最新推荐文章于 2021-07-30 10:52:33 发布

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最新推荐文章于 2021-07-30 10:52:33 发布

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本文介绍了一道经典的算法题目“AvoidTheLakes”，通过深度优先搜索（DFS）的方法来解决农场中最大湖泊面积的问题。文章详细展示了如何利用递归进行网格遍历，并通过实例解释了算法的工作原理。

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Avoid The Lakes

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7956		Accepted: 4183

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

* 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

Sample Output

Source

USACO 2007 November Bronze

这道题只需判断四个方向就行了，找到最大的联通单元

#include <cstdio>  
#include <string.h>    
using namespace std;  
int a[110][110];  
int n,m,k;  
int dfs(int i,int j)  
{  
    if(i<1||j<1||i>n||j>m||a[i][j]==0)  
    return 0;  
    a[i][j]=0;  
    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));    
        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]==1)  
        {  
            int sum=dfs(i,j);  
            if(sum>max)  
            max=sum;  
        }  
        printf("%d\n",max);  
    }  
    return 0;  
}