Poj3620--Avoid The Lakes(基础Dfs)

最新推荐文章于 2018-01-20 12:20:10 发布

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最新推荐文章于 2018-01-20 12:20:10 发布

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文章标签：大数据

原文链接：http://www.cnblogs.com/soTired/p/4702871.html

本文介绍了一个名为AvoidTheLakes的问题，这是一个经典的图搜索问题，通过深度优先搜索（DFS）来找出矩形网格中最大的湖泊（连通区域）大小。湖泊由相邻的水格子组成，相邻定义为上下左右相邻。文章提供了完整的C++代码实现。

Avoid The Lakes

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6908		Accepted: 3680

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

RE: 败给了读题， < 被标记的是湿地 >,

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 using namespace std;
 5 int map[1010][1010], ac[4][2] = {1, 0, -1, 0, 0, 1, 0, -1}, ant;
 6 int Dfs(int x, int y,int m, int n)
 7 {
 8     int nx, ny;
 9     map[x][y] = 0;
10     ant++;
11     for(int i=0; i<4; i++)
12     {
13         nx = x + ac[i][0];
14         ny = y + ac[i][1];
15         if(nx>=1 && nx<=m && ny>=1 && ny <= n && map[nx][ny])
16         {
17                 Dfs(nx, ny, m, n); 
18         }
19     }    
20 } 
21 int main()
22 {
23     int m, n, t;
24     while(~scanf("%d %d %d", &m, &n, &t))
25     {
26         for(int i = 1; i <= m; i++)
27             for(int j = 1; j <= n; j++)
28                 map[i][j] = 0;
29         int a, b;
30         for(int i=1; i<=t; i++)
31         {
32             scanf("%d %d", &a, &b);
33             map[a][b] = 1;    
34         }    
35         int max = 0;
36         for(int i=1; i<=m; i++)
37         {
38             for(int j=1; j<=n; j++)
39             {
40                 if(map[i][j])
41                 {
42                     ant = 0;
43                     Dfs(i, j, m, n);
44                     if(ant > max)
45                         max = ant;
46                 }
47             }
48         }
49         printf("%d\n", max);
50     }
51     return 0;
52 }