POJ 3620 Avoid The Lakes

Avoid The Lakes

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8274 Accepted: 4328

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

3 4 53 22 23 12 31 1

Sample Output

4

题目要求在指定大小的方格栏中寻找有水的最大连通区域，比如题目给出的案例，在一个3*4的区域内被洪水淹没的方格为输入的，可以置该方格为1，置0的为未被淹没的方格。显然最大的连通区域为4，采用搜索来做（dfs bfs都可以）。

1 0 0

0 1 1

1 1 0

0 0 0

#include <iostream>#include <cstdio>using namespace std;bool a[102][102];//默认初始值为0int num;void dfs(int i,int j){    if(a[i][j])    {        num++;        a[i][j]=0;        dfs(i,j+1);        dfs(i,j-1);        dfs(i+1,j);        dfs(i-1,j);    }}int main(){    int r,c,k;    while(scanf("%d %d %d",&r,&c,&k)!=EOF)    {        int x,y,temp;        for(int i=0;i<k;i++)        {            scanf("%d %d",&x,&y);            a[x][y]=1;        }        temp=0;        for(int i=1;i<=r;i++)            for(int j=1;j<=c;j++)            {                if(a[i][j])//寻找矩阵中不是0的情况，即被洪水淹没形成水洼的情况                {                    num=0;                    dfs(i,j);                    if(num>temp)//对每一个节点的dfs产生的连通区域的个数记录并与上一次的进行比较，留下最大的                        temp=num;                }            }        printf("%d\n",temp);    }    return 0;}