2019寒假专题一 dfs求连通块 入门题 B POJ 2386 C POJ 1979

B

题面

Due to recent rains, water has pooled in various places in Farmer John’s field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water (‘W’) or dry land (’.’). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.

Given a diagram of Farmer John’s field, determine how many ponds he has.
Input

• Line 1: Two space-separated integers: N and M

• Lines 2…N+1: M characters per line representing one row of Farmer John’s field. Each character is either ‘W’ or ‘.’. The characters do not have spaces between them.
  Output

• Line 1: The number of ponds in Farmer John’s field.
  Sample Input
  10 12
  W…WW.
  .WWW…WWW
  …WW…WW.
  …WW.
  …W…
  …W…W…
  .W.W…WW.
  W.W.W…W.
  .W.W…W.
  …W…W.
  Sample Output
  3

题面描述

问有多少个水池

分析

求几个连通块，遇到一个W 就将整个水池dfs一遍，记录dfs的次数

代码

#include 
#include
using namespace std;

int n,m; 
char ch[105][105];			
void dfs(int r,int c)
{
	if(r>=n||r<0|| c>=m||c<0) return ;		//3个终止条件 
	else if(ch[r][c]!='W')  return ;
	else ch[r][c]='.';
	
	for(int dr=-1;dr<=1;dr++){
		for(int dc=-1;dc<=1;dc++){
			if(dr||dc) {				//不同时为 0 
				dfs(r+dr,c+dc);
			}
			
		}
	}
	
	return ;			
}



int main()
{
	//int n,m;
	cin>>n>>m;
	int cnt=0;
	
	for(int i=0;i<n;i++) scanf("%s",ch[i]);		//以字符串形式输入每一行 
	 
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++){
			if(ch[i][j]=='W'){ dfs(i,j); cnt++; }	//大写 W 
			
		}
	}
	
	cout<<cnt<<endl;
	
	return 0;
}

C

题面

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can’t move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above.
Input
The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

‘.’ - a black tile
‘#’ - a red tile
‘@’ - a man on a black tile(appears exactly once in a data set)
The end of the input is indicated by a line consisting of two zeros.
Output
For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).
Sample Input
6 9
…#.
…#
…
…
…
…
…
#@…#
.#…#.
11 9
.#…
.#.#######.
.#.#…#.
.#.#.###.#.
.#.#…@#.#.
.#.#####.#.
.#…#.
.#########.
…
11 6
…#…#…#…
…#…#…#…
…#…#…###
…#…#…#@.
…#…#…#…
…#…#…#…
7 7
…#.#…
…#.#…
###.###
…@…
###.###
…#.#…
…#.#…
0 0
Sample Output
45
59
6
13

题面描述

问能到达的黑砖数

分析

找到人站的位置后，分别向上下左右dfs，同时标记走过的黑砖，每走到一块黑砖，数量+1

代码

#include
#include
using namespace std;

int R,C;
int cnt=0;				//
char ch[25][25];

int dr[4]={-1,1,0,0};			//方向数组 
int dc[4]={0,0,-1,1};

void dfs(int r,int c)
{
	if(r<0||c<0||r>=R||c>=C) return ;
	if(ch[r][c]=='#'||ch[r][c]=='Y') return;//若用！='.'  刚开始是 @ 直接被 return 出来。。
	
	ch[r][c]='Y'; 	// Y标记已走过 
	cnt++;			
	
	
	for(int i=0;i<4;i++){
		dfs(r+dr[i],c+dc[i]);
	}
	
	return ;
}


int main()
{
	while(cin>>C>>R , C||R){
		
		cnt=0;									//清空 
		for(int i=0;i<R;i++) scanf("%s",ch[i]);
		
		for(int i=0;i<R;i++){
			for(int j=0;j<C;j++){
				if(ch[i][j]=='@') { dfs(i,j); break;	
				}
				
			}
		}		
		
		
		cout<<cnt<<endl;
	}
	
	return 0;
 }