本文主要是介绍POJ 3620 Avoid The Lakes（连通分量，DFS），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 7825 | Accepted: 4131 |
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
题意：
这英文写的，翻译的。。。。。有道都救不了！
给你一个 N*M 的矩形，以及 K 个坐标点作为淹没的标记，问你最大的淹没区域是多少。
思路：
利用 DFS 计算每个连通分量的大小比较后取最大值。
代码：
#include<stdio.h>
#include<string.h>
#define MYDD 1103//N (1 ≤ N ≤ 100) rows and M (1 ≤ M ≤ 100) columns 行，列
// either dry or submerged 干燥或者淹没
int N,M,K;
int map[128][128];//记录淹没的坐标点
int sum;//遍历当前的淹没面积
int max_sum;//最大的淹没面积
int dx[]= {0,0,-1,1};
int dy[]= {-1,1,0,0};//移动的四个方向 void DFS(int x,int y) {sum++;if(!map[x][y])//该点不被淹没 return ;map[x][y]=0;//标记访问，即干燥 for(int j=0; j<4; j++) {int gx=x+dx[j];int gy=y+dy[j];if(gx>=1&&gx<=N&&gy>=1&&gy<=M&&map[gx][gy])DFS(gx,gy);}
}int main() {while(scanf("%d%d%d",&N,&M,&K)!=EOF) {max_sum=0;memset(map,0,sizeof(map));int row,column;for(int j=0; j<K; j++) {scanf("%d%d",&row,&column);map[row][column]=1;}for(int j=1; j<=N; j++) {for(int k=1; k<=M; k++) {if(map[j][k]) {sum=0;DFS(j,k);if(sum>max_sum)max_sum=sum;}}}printf("%d\n",max_sum);}return 0;
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