uva 816 Ancient Messages (BFS)

其实就是BFS求最短路，先说明为什么BFS可以求最短路。因为BFS是广度搜索，是一层一层的，最先到达约定条件的那一层一定是最短的。

本题复杂在方向判断上，首先需要用数组has_edge[r][c][dir][trun]来记录从方向dir到达r,c点，转向turn是否可行。

然后用d[r][c][dir]来记录从方向dir到r,c的长度，p[r][c][dir]记录从方向dir到r,c的父节点。

关于转向，可以在每个点都用一个for(int i=0;i<3;i++),当i==0表示直线,i==1表示左转，i==2表示右转。对每个方向都判断一下，因为我们用has_edge数组记录在点r,c的某个转向是否可行。

import java.util.LinkedList;import java.util.Queue;import java.util.Scanner;import java.util.Vector;public class Main {static String dirs = "NESW";static String turns = "FLR";static int[] dr = {-1,0,1,0};static int[] dc = {0,1,0,-1};static int r0,r1,r2,c0,c1,c2,dir;static int[][][][] has_edge = new int[10][10][4][3];static int[][][] d = new int[10][10][4];static Node[][][] p = new Node[10][10][4];public static void main(String[] args) {Scanner scan = new Scanner(System.in);while(true){String name = scan.next();if("END".equals(name))break;for(int i=1;i<=9;i++){for(int j=1;j<=9;j++){for(int k=0;k<4;k++){for(int l=0;l<3;l++){has_edge[i][j][k][l] = 0;}}}}r0 = scan.nextInt();c0 = scan.nextInt();dir = dir_id(scan.next().charAt(0));r2 = scan.nextInt();c2 = scan.nextInt();r1 = r0+dr[dir];c1 = c0+dc[dir];while(true){int r = scan.nextInt();if(r==0)break;int c = scan.nextInt();while(true){String str = scan.next();char dd = str.charAt(0);if(dd=='*')break;for(int i=1;i<str.length();i++){has_edge[r][c][dir_id(dd)][turn_id(str.charAt(i))]=1;}}}System.out.println(name);solve();}}public static void solve(){for(int i=1;i<=9;i++){for(int j=1;j<=9;j++){for(int k=0;k<4;k++){d[i][j][k] = 0;p[i][j][k] = null;}}}Queue<Node> q = new LinkedList<>();Node u = new Node(r1,c1,dir);d[u.r][u.c][u.dir] = 0;q.add(u);while(!q.isEmpty()){u = q.poll();if(u.r==r2&&u.c==c2){print_ans(u);return;}else{for(int i=0;i<3;i++){Node v = walk(u,i);if(has_edge[u.r][u.c][u.dir][i]==1&&inside(v.r,v.c)&&d[v.r][v.c][v.dir]==0){d[v.r][v.c][v.dir] = d[u.r][u.c][u.dir]+1;p[v.r][v.c][v.dir] = u;q.add(v);}}}}System.out.println("  No Solution Possible");}public static void print_ans(Node u){Vector<Node> v = new Vector<>();for(;;){v.add(u);if(d[u.r][u.c][u.dir]==0)break;u = p[u.r][u.c][u.dir];}v.add(new Node(r0,c0,dir));int cnt = 0;for(int i=v.size()-1;i>=0;i--){if(cnt%10==0)System.out.print(" ");System.out.printf(" (%d,%d)",v.get(i).r,v.get(i).c);if(++cnt%10==0)System.out.println();}if(v.size()%10!=0){System.out.println();}}public static boolean inside(int r,int c){return r>=1&&r<=9&&c>=1&&c<=9;}public static Node walk(Node u,int turn){int dir = u.dir;if(turn==1)dir = (dir+3)%4;if(turn==2)dir = (dir+1)%4;return new Node(u.r+dr[dir],u.c+dc[dir],dir);}public static int dir_id(char c){return dirs.indexOf(c);}public static int turn_id(char c){return turns.indexOf(c);}static class Node{int r,c,dir;public Node(int r,int c,int dir){this.r = r;this.c = c;this.dir = dir;}}}