POJ3620

Avoid The Lakes

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6697 Accepted: 3595

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

这道题其实直接用深度搜索就可以解决，以下是我的代码:
#include<iostream>#include<cstdio>#include<memory.h>#define max(a,b) ((a>b)?(a):(b))using namespace std;const int Max=101;int MAX,Num;int a[Max][Max],d[Max][Max];int N,M,K;int exam(int x,int y){int X[2]={-1,1};int Y[2]={-1,1};for(int j=0;j<2;j++){   if(d[x][y+Y[j]]==0&&a[x][y+Y[j]]==1){   d[x][y+Y[j]]=1;        Num++;   exam(x,y+Y[j]);   }     }   for(int j=0;j<2;j++){   if(d[x+X[j]][y]==0&&a[x+X[j]][y]==1){   d[x+X[j]][y]=1;       Num++;   exam(x+X[j],y);   }    }    return 0;}int main(){int b[Max],c[Max];while(scanf("%d%d%d",&N,&M,&K)!=EOF){memset(a,0,sizeof(a));memset(d,0,sizeof(d));MAX=0;for(int i=0;i<K;i++){int m,n;scanf("%d%d",&m,&n);b[i]=m-1;c[i]=n-1;a[m-1][n-1]=1;}for(int j=0;j<K;j++){Num=1;d[b[j]][c[j]]=1;exam(b[j],c[j]);        MAX=max(MAX,Num);memset(d,0,sizeof(d));}printf("%d\n",MAX);}return 0;}

不同于一般的深度搜索，这里我采用的是全局变量来存储符合条件的点的个数，并且用了同样大的数组来存储每次选点的情况，可能相比其他的解题方法来说要麻烦一点。

在这个程序中其实还有点问题，如果大家看出来了请给我留言，谢谢。