uva 1600 Patrol Robot (草地机器人)BFS

这道题一开始真的没思路，直接dfs，结果WA了；

看了题解才知道用bfs，这道题使我加深了对BFS的理解。

这道题就是在普通的BFS上加上了一个额外条件，障碍物，所以只需要记录经过每个点时所穿过的障碍物，比如，定义layer表示穿过障碍物的个数，u表示上一个位置，则如果当前位置是1,则layer = u.layer+1,否则layer为0。如果layer<=k，并且这个点的第layer层也没有访问过,则将这个点加入到队列中，详情看代码

TLE代码:

import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

public class Main {

static int m,n,k;
static int maxn = 25;
static int[][] G = new int[maxn+1][maxn+1];
static int[][] vis = new int[maxn+1][maxn+1];
static int[] dr = {-1,0,1,0};
static int[] dc = {0,1,0,-1};
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int t = scan.nextInt();
while (t--!= 0) {
m = scan.nextInt();
n = scan.nextInt();
k = scan.nextInt();
for(int i=0;i<=maxn;i++){
for(int j=0;j<=maxn;j++){
vis[i][j] = 0;
G[i][j] = 0;
}
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
G[i][j] = scan.nextInt();
}
}
System.out.println(bfs());
}
}

public static int bfs() {
Queue<Node> q = new LinkedList<>();
q.add(new Node(0, 0, 1, 0));
vis[0][0] = 1;
while (!q.isEmpty()) {
Node u = q.peek();
q.poll();
if (u.r == m - 1 && u.c == n - 1) {
return u.d - 1;
}
for (int i = 0; i < 4; i++) {
int r = u.r + dr[i];
int c = u.c + dc[i];
if (r >= 0 && r < m && c >= 0 && c < n) {
if (G[r][c] == 0) {
q.add(new Node(r, c, u.d + 1, 0));
} else {
if ((u.layer + 1) <= k) {
q.add(new Node(r, c, u.d + 1, u.layer + 1));
}
}
}
}
}
return -1;
}

static class Node{
int r,c,d,layer;
public Node(int r,int c,int d,int layer){
this.r = r;
this.c = c;
this.d = d;
this.layer = layer;
}
}
}

AC代码:

import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

public class Main {

static int m,n,k;
static int maxn = 25;
static int[][] G = new int[maxn+1][maxn+1];
static int[][][] vis = new int[maxn+1][maxn+1][maxn+1];
static int[] dr = {-1,0,1,0};
static int[] dc = {0,1,0,-1};
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int t = scan.nextInt();
while (t--!= 0) {
m = scan.nextInt();
n = scan.nextInt();
k = scan.nextInt();
for(int i=0;i<=maxn;i++){
for(int j=0;j<=maxn;j++){
for(int k=0;k<=maxn;k++){
vis[i][j][k] = 0;
}
G[i][j] = 0;
}
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
G[i][j] = scan.nextInt();
}
}
System.out.println(bfs());
}
}

public static int bfs() {
Queue<Node> q = new LinkedList<>();
q.add(new Node(0, 0, 1, 0));
vis[0][0][0] = 1;
while (!q.isEmpty()) {
Node u = q.peek();
q.poll();
if (u.r == m - 1 && u.c == n - 1) {
return u.d - 1;
}
for (int i = 0; i < 4; i++) {
int r = u.r + dr[i];
int c = u.c + dc[i];
int layer = u.layer;
if (r >= 0 && r < m && c >= 0 && c < n) {
if (G[r][c] == 0) {
layer=0;
} else {
layer++;
}

if(layer<=k&&vis[r][c][layer]==0){//加上vis[r][c][layer]的判断是为了避免超时，如果去掉，一定超时。
q.add(new Node(r,c,u.d+1,layer));
}
vis[r][c][layer] = 1;
}
}
}
return -1;
}

static class Node{
int r,c,d,layer;
public Node(int r,int c,int d,int layer){
this.r = r;
this.c = c;
this.d = d;
this.layer = layer;
}
}
}