CSP 201312-5 I‘m stuck Java

题目并不难，其中一个解决思路，就是把从起点能够遍历到的位置标记上，然后再把从终点把能够遍历的位置标记上，之后筛选出满足题目要求的个数即可。

import java.util.*;

public class Main{
    
    //代表输入的字符
    static char[][] g;
    //偏移量
    static int[] dx = {-1, 0, 1, 0};
    static int[] dy = {0, 1, 0, -1};
    //从起点能到达的点的标识
    static boolean[][] vis1;
    //能够到达终点的点的标识
    static boolean[][] vis2;
    
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        
        int R = input.nextInt();
        int C = input.nextInt();
        
        g = new char[R][C];
        vis1 = new boolean[R][C];
        vis2 = new boolean[R][C];
        for (int i = 0; i < R; i++) {
            String str = input.next();
            for (int j = 0; j < C; j++) {
                g[i][j] = str.charAt(j);
            }
        }
        
        int x = -1;
        int y = -1;
        for (int i = 0; i < R; i ++) {
            for (int j = 0; j < C; j ++) {
                if (g[i][j] == 'S') {
                    dfs1(i, j);
                } else if (g[i][j] == 'T') {
                    dfs2(i, j);
                    x = i;
                    y = j;
                }
            }
        }

        
        if (!vis1[x][y]) {
            System.out.println("I'm stuck!");
            System.exit(0);
        }
        
        int res = 0;
        for (int i = 0; i < R; i ++) {
            for (int j = 0; j < C; j ++) {
                if (vis1[i][j] && !vis2[i][j]) {
                    res++;
                }
            }
        }
        
        System.out.println(res);
        
        input.close();
    }
    
    //标记起点能够到达的节点
    private static void dfs1(int x, int y) {
        vis1[x][y] = true;
        for (int i = 0; i < 4; i++) {
            int r = x + dx[i];
            int c = y + dy[i];
            
            if (r < 0 || c < 0 || r >= g.length || c >= g[0].length || g[r][c] == '#' || vis1[r][c]) {
                continue;
            }
            
            //检测从(x,y)能否到达(r,c)
            if (check(x, y, i)) {
                dfs1(r, c);
            }
        }
     
    }
    
    //标记重点能够到达的节点
    private static void dfs2(int x, int y) {
        vis2[x][y] = true;
        for (int i = 0; i < 4; i++) {
            int r = x + dx[i];
            int c = y + dy[i];
            if (r < 0 || c < 0 || r >= g.length || c >= g[0].length || g[r][c] == '#' || vis2[r][c]){
                continue;
            }
            //检测能否通过(r,c)到达(x,y)
            if (check(r, c, i ^ 2)) {
                dfs2(r, c);
            }
        }
    }
    
    private static boolean check(int x, int y, int w) {
        char c = g[x][y];
        
        if (c == 'S' || c == '+' || c == 'T') {
            return true;
        } else if (c == '|' && (w == 0 || w == 2)) {
            return true;
        } else if (c == '-' && (w == 1 || w == 3)) {
            return true;
        } else if (c == '.' && w == 2) {
            return true;
        } else {
            return false;
        }
    }
}