POJ 3620 Avoid The Lakes

原题传送门：http://poj.org/problem?id=3620

 

Avoid The Lakes

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6211		Accepted: 3370

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 5
3 2
2 2
3 1
2 3
1 1

Sample Output

4

Source

USACO 2007 November Bronze

 

 

分析：这道其实是深搜的水题啊！可是因为while(scanf("%d%d%d",&n,&m,&k) == 3)我没写==3这个条件结果一直给我超时orz。

 

//深度优先遍历 DFS  POJ 3602
#include<stdio.h>
#include<stack>
#include<string.h>
#include<cmath>
using namespace std;
int N,M,K,a,b;
const int MAXN = 110;
int form[MAXN][MAXN];
bool visit[MAXN][MAXN];
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
int count;
int ans;
void DFS(int x,int y){
    int xx,yy;

    for(int i=0;i<4;i++){
        xx = x + dir[i][0];
        yy = y + dir[i][1];

        if(xx<1||yy<1||xx>N||yy>M)
            continue;
        if(form[xx][yy] && !visit[xx][yy]){
            count++;
            visit[xx][yy] = 1;
            DFS(xx,yy);
        }
    }
}
int main(){
    while(scanf("%d%d%d",&N,&M,&K)==3){
        ans = 0;
        memset(form,0,sizeof(form));
        memset(form,false,sizeof(visit));
        while(K--){
            scanf("%d%d",&a,&b);
            form[a][b] = 1;
        }
        for(int i=1;i<=N;i++){
            for(int j=1;j<=M;j++){
                if(form[i][j]&&!visit[i][j]){
                        count = 1;
                        visit[i][j]=1;
                        DFS(i,j);
                        ans = max(ans,count);
                }
            }
        }

        printf("%d\n",ans);

    }
    return 0;



}