[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];
	}
}