[输入输出]:
[解题分析]:
题解：双重bfs，先对箱子bfs，然后判断这种bfs是否可达（对人bfs）
下面是AC过的代码：
import java.util.*;
import java.io.*;
public class Main{
int r;//地图行数
int c;//地图列数
int begx, begy;//箱子开始坐标
int endx, endy;//目标坐标
int begsx, begsy;//人开始坐标
char map[][];//地图
int[] dx ={-1, 1, 0, 0};//人和箱子都有四个方向可移动
int[] dy ={0, 0, 1, -1};
char[] P ={'N', 'S', 'E', 'W'};//表示箱子向某个方向移动
char[] M ={'n', 's', 'e', 'w'};//表示人向某个方向移动
Node f=new Node(0,0,0,0,"");
Node g=new Node(0,0,0,0,"");
node1 F=new node1(0,0,"");
node1 G=new node1(0,0,"");
int mark[][];//标志数组，表示地图上某一位置mark[i][j]是否访问过。
public Main(char[][] map,int r,int c,int begx,int begy,int endx,int endy,int begsx,int begsy){
this.map=map;
this.r=r;
this.c=c;
this.begx=begx;
this.begy=begy;
this.endx=endx;
this.endy=endy;
this.begsx=begsx;
this.begsy=begsy;
mark=new int[r][c];
}
public boolean ok(int x,int y) {
if (x >= 0 && x < r && y >= 0 && y < c) return true;
return false;
}
public boolean SToB(int bx,int by,int ex, int ey) {//人到箱子BFS
int[][] Mark1= new int[r][c]; //标志数组，表示地图上某一位置Mark1[i][j]是否访问过。
Queue P = new LinkedList();
Mark1[bx][by] = 1;
Mark1[f.bx][f.by] = 1;
F.x = bx; F.y = by;
F.ans = "";
if (bx == ex && by == ey) return true;//到达目标
P.offer(new node1(F.x,F.y,F.ans));
while (!P.isEmpty()) {
F = P.poll();
for (int i = 0; i < 4; ++i) {//向四个方向扩展
G.x = F.x + dx[i];
G.y = F.y + dy[i];
if (!ok(G.x, G.y) || map[G.x][G.y] == '#') continue;
if (Mark1[G.x][G.y]==0) {//此点没有访问过
Mark1[G.x][G.y] = 1;//标记为已访问
G.ans = F.ans + M[i];
if (G.x == ex && G.y == ey) {
F = G;
return true;
}
P.add(new node1(G.x,G.y,G.ans));
}
}
}
return false;
}
public boolean bfs() {//箱子向目标bfs
Queue Q = new LinkedList();
f.bx = begx; f.by = begy;//f:人与箱子所在的位置
f.px = begsx; f.py = begsy;
f.ans = "";
mark[begx][begy] = 1;
Q.offer(new Node(f.bx,f.by,f.px,f.py,f.ans));
while (!Q.isEmpty()) {
f = Q.poll();//出队列
for (int i = 0; i < 4; ++i) {//检查f的所有邻接点，向四个方向扩展
int newx = f.bx + dx[i];
int newy = f.by + dy[i];//箱子前一位置
int tx = f.bx - dx[i];
int ty = f.by - dy[i];//箱子后一位置
if (!ok(newx, newy) || map[newx][newy] == '#' || !ok(tx, ty)
|| map[tx][ty] == '#' || mark[newx][newy]==1) continue;
if (SToB(f.px, f.py, tx, ty)) {//检查人能否来
g.bx = newx; g.by = newy;
g.px = f.bx; g.py = f.by;
g.ans = f.ans + F.ans + P[i];
if (g.bx == endx && g.by == endy)
{
return true;
}
mark[newx][newy] = 1;
Q.add(new Node(g.bx,g.by,g.px,g.py,g.ans));
}
}
}
return false;
}
public static void main(String args[]) {
Scanner in=new Scanner(System.in);
int r=0;
int c=0;
String s="";
int begx=0;
int begy=0;
int endx=0;
int endy=0;
int begsx=0;
int begsy=0;
char[][] map=null;
int t=1;
while(in.hasNext()){
r=in.nextInt();
c=in.nextInt();
if(r==0&&c==0) break;
map=new char[r][c];
for(int i = 0; i
- 1475Main.rar (1.8 KB)
- 下载次数: 13
- 查看图片附件
