bash
day02 图论part02
今日任务:岛屿数量,岛屿的最大面积
都是一个模子套出来的
https://programmercarl.com/kamacoder/0099.岛屿的数量深搜.html#思路
往日任务:
day01 图论part01
今日任务:图论理论基础/所有可到达的路径
代码随想录图论视频部分还没更新
https://programmercarl.com/kamacoder/图论理论基础.html#图的基本概念day02
岛屿数量
dfs
import java.util.Scanner;
public class Main{
public static int[][] dir ={{0,1},{1,0},{-1,0},{0,-1}};
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int m = sc.nextInt();
int n = sc.nextInt();
int[][] grid = new int[m][n];
for(int i = 0 ; i < m; i++){
for(int j = 0; j < n; j++){
grid[i][j] = sc.nextInt();
}
}
boolean[][] visited = new boolean[m][n];
int count = 0;
for(int i = 0 ; i < m; i++){
for(int j = 0; j < n; j++){
if( visited[i][j] == false && grid[i][j] == 1){
count++;
visited[i][j] = true;//访问过了
dfs(grid,i,j,visited);//一直找临近陆地直到找不到
}
}
}
System.out.println(count);
}
private static void dfs(int[][] grid,int x,int y,boolean[][] visited){
for(int i = 0; i < 4; i++){
//x += dir[i][0];//这里错了,x和y需要用四次,可是刚用一次值就改变了
//y += dir[i][1];
int x1 = x + dir[i][0];
int y1 = y + dir[i][1];
if(x1<0||y1<0||x1>= grid.length||y1>=grid[0].length)
continue;//越界则继续判断下一个旁边的位置
if(!visited[x1][y1] && grid[x1][y1]==1)//旁边是没遇到过的陆地
{
visited[x1][y1]=true;
dfs(grid,x1,y1,visited);//继续找临近陆地
}
}
}
}bfs
main方法一样,dfs和bfs有细微的差别,dfs是遇到陆地就递归直到越界,bfs是遇到陆地就加到queue里面直到queue为空
linkedlist实现queue,用到了isEmpty方法,peek方法和poll方法
import java.util.*;
public class Main {
public static int[][] dir = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};//下右上左逆时针遍历
public static void bfs(int[][] grid, boolean[][] visited, int x, int y) {
Queue<pair> queue = new LinkedList<pair>();//定义坐标队列，没有现成的pair类，在下面自定义了
queue.add(new pair(x, y));//第一个位置入队
visited[x][y] = true;//遇到入队直接标记为优先，
// 否则出队时才标记的话会导致重复访问，比如下方节点会在右下顺序的时候被第二次访问入队
while (!queue.isEmpty()) {
int curX = queue.peek().first;
int curY = queue.poll().second;//当前横纵坐标
for (int i = 0; i < 4; i++) {
//顺时针遍历新节点next，下面记录坐标
int nextX = curX + dir[i][0];
int nextY = curY + dir[i][1];
if (nextX < 0 || nextX >= grid.length || nextY < 0 || nextY >= grid[0].length) {
continue;
}//去除越界部分
if (!visited[nextX][nextY] && grid[nextX][nextY] == 1) {
queue.add(new pair(nextX, nextY));
visited[nextX][nextY] = true;//逻辑同上
}
}
}
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int m = sc.nextInt();
int n = sc.nextInt();
int[][] grid = new int[m][n];
boolean[][] visited = new boolean[m][n];
int ans = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
grid[i][j] = sc.nextInt();
}
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (!visited[i][j] && grid[i][j] == 1) {
ans++;
bfs(grid, visited, i, j);
}
}
}
System.out.println(ans);
}
// 定义 pair 类来表示坐标
public static class pair {
int first; // 横坐标
int second; // 纵坐标
// 构造函数
public pair(int x, int y) {
this.first = x;
this.second = y;
}
}
}岛屿的最大面积
dfs
套岛屿数量的模板,变化很少(话说这道题怎么没有答案啊)
import java.util.Scanner;
public class Main{
public static int count;//这里变了
public static int[][] dir ={{0,1},{1,0},{-1,0},{0,-1}};
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int m = sc.nextInt();
int n = sc.nextInt();
int[][] grid = new int[m][n];
for(int i = 0 ; i < m; i++){
for(int j = 0; j < n; j++){
grid[i][j] = sc.nextInt();
}
}
boolean[][] visited = new boolean[m][n];
int result = 0;//这里变了
for(int i = 0 ; i < m; i++){
for(int j = 0; j < n; j++){
if( visited[i][j] == false && grid[i][j] == 1){
count = 1;
visited[i][j] = true;
dfs(grid,i,j,visited);
result = Math.max(result, count);//这里变了
}
}
}
System.out.println(result);//这里变了
}
private static void dfs(int[][] grid,int x,int y,boolean[][] visited){
for(int i = 0; i < 4; i++){
int x1 = x + dir[i][0];
int y1 = y + dir[i][1];
if(x1<0||y1<0||x1>= grid.length||y1>=grid[0].length)
continue;
if(!visited[x1][y1] && grid[x1][y1]==1)
{
visited[x1][y1]=true;
count++;//这里变了
dfs(grid,x1,y1,visited);
}
}
}
}bfs
套模版,基本没变化
import java.util.*;
public class Main {
public static int count;//多了这一行
public static int[][] dir = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
public static void bfs(int[][] grid, boolean[][] visited, int x, int y) {
Queue<pair> queue = new LinkedList<pair>();
queue.add(new pair(x, y));
visited[x][y] = true;
while (!queue.isEmpty()) {
int curX = queue.peek().first;
int curY = queue.poll().second;
for (int i = 0; i < 4; i++) {
int nextX = curX + dir[i][0];
int nextY = curY + dir[i][1];
if (nextX < 0 || nextX >= grid.length || nextY < 0 || nextY >= grid[0].length) {
continue;
}
if (!visited[nextX][nextY] && grid[nextX][nextY] == 1) {
queue.add(new pair(nextX, nextY));
visited[nextX][nextY] = true;
count++;//多了这一行
}
}
}
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int m = sc.nextInt();
int n = sc.nextInt();
int[][] grid = new int[m][n];
boolean[][] visited = new boolean[m][n];
int result = 0;//这里
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
grid[i][j] = sc.nextInt();
}
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (!visited[i][j] && grid[i][j] == 1) {
count = 1;
bfs(grid, visited, i, j);
result = Math.max(result, count);//这里
}
}
}
System.out.println(result);
}
public static class pair {
int first;
int second;
public pair(int x, int y) {
this.first = x;
this.second = y;
}
}
}感谢大佬分享: