poj2386:Lake Counting
作者:互联网
给定一个图,'.'表示陆地,'W'表示有水,求水洼的数量
遍历图,若遍历到'W'则进行深搜,将该点置为'.',遍历该点所有邻接点,若为'W'则继续深搜
在遍历图的过程中进行调用dfs的次数即为水洼数(怎么感觉像求连通分量......)
#include<stdio.h>
#include<iostream>
using namespace std;
const int MAXSIZE = 100 + 10;
char filed[MAXSIZE][MAXSIZE];
int n, m;
void dfs(int x, int y){
filed[x][y] = '.'; //将(x,y)置为 '.'
//遍历8个方向
for(int dx = -1; dx <= 1; ++dx){
for(int dy = -1; dy <= 1; ++dy){
//(nx,ny)为下一个邻接点
int nx = x + dx;
int ny = y + dy;
//判定邻接点是否在范围内 且 如果是'W'则深搜
if( (nx >= 0 && nx < n) && (ny >= 0 && ny < m) && filed[nx][ny] == 'W'){
dfs(nx, ny);
}
}
}
}
int main(){
scanf("%d%d", &n, &m);
for(int i = 0; i < n; ++i){
getchar();
for(int j = 0; j < m; ++j){
scanf("%c", &filed[i][j]);
}
}
int answer = 0;
for(int i = 0; i < n; ++i){
for(int j = 0; j < m; ++j){
if(filed[i][j] == 'W'){
dfs(i, j);
++answer; //总共进行深搜的次数即为所求
}
}
}
printf("%d\n", answer);
return 0;
}
标签:遍历,ny,int,Lake,++,&&,filed,Counting,poj2386 来源: https://www.cnblogs.com/dctwan/p/15881782.html