101.孤岛的总面积
基础题目 可以自己尝试做一做 。
方法1:深度搜索
import java.util.*;
public class Main{
public static int[][] dirc = {
{0,1},{1,0},{0,-1},{-1,0}};
public static void main (String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int m = scanner.nextInt();
int[][] graph = new int[n][m];
for(int i = 0; i < n ; i++){
for(int j = 0; j < m; j++){
graph[i][j] = scanner.nextInt();
}
}
boolean[][] visited = new boolean[n][m];
for(int i = 0; i < n; i++){
if(graph[i][0] == 1){
dfs(graph, visited, i , 0);
}
if(graph[i][m-1] == 1){
dfs(graph, visited, i , m-1);
}
}
for(int j = 0; j < m; j++){
if(graph[0][j] == 1){
dfs(graph, visited, 0, j);
}
if(graph[n-1][j] == 1){
dfs(graph, visited, n-1, j);
}
}
int sum = 0;
for(int i = 0; i < n; i++){
for(int j = 0; j < m; j++){
if(graph[i][j] == 1){
sum ++;
}
}
}
System.out.println(sum);
}
public static void dfs(int[][] graph, boolean[][] visited, int x, int y){
if(!(x >= 0 && x < graph.length && y >= 0 && y < graph[0].length )){
return ;
}
if(visited[x][y] || graph[x][y] == 0){
return ;
}
visited[x][y] = true;
graph[x][y] = 0;
for(int i = 0; i < dirc.length; i++){
int nextX = x + dirc[i][0];
int nextY = y + dirc[i][1];
dfs(graph, visited, nextX, nextY);
}
}
}
方法2:广度搜索
import java.util.*;
public class Main{
public static int[][] dirc = {
{0,1},{1,0},{0,-1},{-1,0}};
public static void main (String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int m = scanner.nextInt();
int[][] graph = new int[n][m];
for(int i = 0; i < n ; i++){
for(int j = 0; j < m; j++){
graph[i][j] = scanner.nextInt();
}
}
boolean[][] visited = new boolean[n][m];
for(int i = 0; i < n; i++){
if(graph[i][0] == 1){
bfs(graph, visited, i , 0);
}
if(graph[i][m-1] == 1){
bfs(graph, visited, i , m-1);
}
}
for(int j = 0; j < m; j++){
if(graph[0][j] == 1){
bfs(graph, visited, 0, j);
}
if(graph[n-1][j] == 1){
bfs(graph, visited, n-1, j);
}
}
int sum = 0;
for(int i = 0; i < n; i+&