[topcoder]AvoidRoads

二维动态规划。和某一道leetcode的题目差不多。就是多了blocks的数组或集合。

本次解题的心得有：1.根据题意使用集合表示阻碍；2.使用字符串的形式表示整数的pair，简洁明了；3.p1到p2的阻碍其实是双向的；4.可以不用首行首列的全0；5.mx[i][j]和mx[i-1][j]和mx[i-1][j]可以分别加的。

import java.util.*;public class AvoidRoads
{public long numWays(int width, int height, String[] bad) {HashMap<String,HashSet<String>> blocks = new HashMap<String,HashSet<String>>();for (String badStr : bad) {String[] bl = badStr.split(" ");int x1 = Integer.parseInt(bl[0]);int y1 = Integer.parseInt(bl[1]);int x2 = Integer.parseInt(bl[2]);int y2 = Integer.parseInt(bl[3]);String p1 = "" + x1+ ":" + y1;String p2 = "" + x2 + ":" + y2;// p1 -> p2 && p2-> p1 are blockedif (!blocks.containsKey(p1)) {HashSet<String> set = new HashSet<String>();blocks.put(p1, set);}if (!blocks.containsKey(p2)) {HashSet<String> set = new HashSet<String>();blocks.put(p2, set);}blocks.get(p1).add(p2);blocks.get(p2).add(p1);}long mx[][] = new long[width+1][height+1];for (int i = 0; i < width+1; i++) {for (int j = 0; j < height+1; j++) {if (i == 0 && j == 0) {mx[i][j] = 1;}else {String s0 = ""+i+":"+j;String s1 = ""+(i-1)+":"+j;String s2 = ""+i+":"+(j-1);if (i > 0 && !(blocks.containsKey(s1) && blocks.get(s1).contains(s0))) {mx[i][j] += mx[i-1][j];}if (j > 0 && !(blocks.containsKey(s2) && blocks.get(s2).contains(s0))) {mx[i][j] += mx[i][j-1];}}}}return mx[width][height];}
}

转载于:https://www.cnblogs.com/lautsie/p/3258732.html

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