[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 blocked

			if (!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];

	}

}