skiing
1 package noj_skiing; 2 3 import java.util.*; 4 import java.math.*; 5 6 public class Main { 7 8 public static void main(String[] args) { 9 Solution s = new Solution(); 10 Scanner sc = new Scanner(System.in); 11 int N = sc.nextInt(); 12 int[] result = new int[N]; 13 for(int i = 0; i < N; i++) { 14 int rowNum = sc.nextInt(); 15 int colNum = sc.nextInt(); 16 int[][] info = new int[rowNum][colNum]; 17 for(int r = 0; r < rowNum; r++) { 18 for(int c = 0; c < colNum; c++) { 19 info[r][c] = sc.nextInt(); 20 } 21 } 22 result[i] = s.getResult(info, rowNum, colNum); 23 } 24 sc.close(); 25 for(int i : result) { 26 System.out.println(i); 27 } 28 } 29 30 } 31 class Solution { 32 public int getResult(int[][] info, int rowNum, int colNum) { 33 int max = 0; 34 int[][] states = new int[rowNum][colNum]; //make sure equal 0 all 35 for(int r = 0; r < rowNum; r++) { 36 for(int c = 0; c < colNum; c++) { 37 max = Math.max(max, getMaxLength(info, r, c, rowNum, colNum, states)); 38 } 39 } 40 return max; 41 } 42 public int getMaxLength(int[][] info, int row, int col, int rowNum, int colNum, int[][] states) { 43 44 if(row < 0 || row >= rowNum || col < 0 || col >= colNum) //如果输入的坐标越界的处理 45 return 0; 46 if(states[row][col] != 0) //所求位置的最大长度求过 47 return states[row][col]; 48 49 int nextStep = findNextStep(info, row, col, rowNum, colNum); 50 if(nextStep == 0) //所求位置在队尾 51 return states[row][col] = 1; 52 int maxLengthUp = (nextStep & 8) == 8 ? getMaxLength(info, row - 1, col, rowNum, colNum, states) : 0; 53 int maxLengthDown = (nextStep & 4) == 4 ? getMaxLength(info, row + 1, col, rowNum, colNum, states) : 0; 54 int maxLengthLeft = (nextStep & 2) == 2 ? getMaxLength(info, row, col - 1, rowNum, colNum, states) : 0; 55 int maxLengthRight = (nextStep & 1) == 1 ? getMaxLength(info, row, col + 1, rowNum, colNum, states) : 0; 56 57 return max_4(maxLengthUp, maxLengthDown, maxLengthLeft, maxLengthRight) + 1; 58 } 59 public int findNextStep(int[][] info, int row, int col, int rowNum, int colNum) { 60 int result = 0; 61 62 if(row != 0 && info[row - 1][col] < info[row][col]) //up 63 result |= 8; 64 if(row != rowNum - 1 && info[row + 1][col] < info[row][col]) //down 65 result |= 4; 66 if(col != 0 && info[row][col - 1] < info[row][col]) //left 67 result |= 2; 68 if(col != colNum - 1 && info[row][col + 1] < info[row][col]) //right 69 result |= 1; 70 71 return result; 72 } 73 public int max_4(int i_1, int i_2, int i_3, int i_4) { 74 return Math.max(Math.max(i_1, i_2), Math.max(i_3, i_4)); 75 } 76 }
感觉不是很难，思路一上来就可以想出来，具体一些细节的处理也没有很坑的地方。
