DP21 LIS的变种问题-建桥问题 Variations of LIS-Building Bridge @geeksforgeeks

We have discussed Dynamic Programming solution for Longest Increasing Subsequence problem inthis post and a O(nLogn) solution in this post. Following are commonly asked variations of the standard LIS problem.

1. Building Bridges: Consider a 2-D map with a horizontal river passing through its center. There are n cities on the southern bank with x-coordinates a(1) … a(n) and n cities on the northern bank with x-coordinates b(1) … b(n). You want to connect as many north-south pairs of cities as possible with bridges such that no two bridges cross. When connecting cities, you can only connect city i on the northern bank to city i on the southern bank.

8     1     4     3     5     2     6     7  
<---- Cities on the other bank of river---->
--------------------------------------------
  <--------------- River--------------->
--------------------------------------------
1     2     3     4     5     6     7     8
<------- Cities on one bank of river------->

Source: Dynamic Programming Practice Problems. The link also has well explained solution for the problem.

2. Maximum Sum Increasing Subsequence: Given an array of n positive integers. Write a program to find the maximum sum subsequence of the given array such that the intgers in the subsequence are sorted in increasing order. For example, if input is {1, 101, 2, 3, 100, 4, 5}, then output should be {1, 2, 3, 100}. The solution to this problem has been published here.

3. The Longest Chain You are given pairs of numbers. In a pair, the first number is smaller with respect to the second number. Suppose you have two sets (a, b) and (c, d), the second set can follow the first set if b < c. So you can form a long chain in the similar fashion. Find the longest chain which can be formed. The solution to this problem has been published here.

4. Box Stacking You are given a set of n types of rectangular 3-D boxes, where the i^th box has height h(i), width w(i) and depth d(i) (all real numbers). You want to create a stack of boxes which is as tall as possible, but you can only stack a box on top of another box if the dimensions of the 2-D base of the lower box are each strictly larger than those of the 2-D base of the higher box. Of course, you can rotate a box so that any side functions as its base. It is also allowable to use multiple instances of the same type of box.
Source: Dynamic Programming Practice Problems. The link also has well explained solution for the problem.

对于二维的问题，先把一维固定住（排序），在用动态规划处理剩下一维。

package DP;

import java.util.Arrays;
import java.util.Comparator;

public class BuildingBridges {

	public static void main(String[] args) {
		Pair[] A = new Pair[7];
		A[0] = new Pair(22,4);
		A[1] = new Pair(2,6);
		A[2] = new Pair(10,3);
		A[3] = new Pair(15,12);
		A[4] = new Pair(9,8);
		A[5] = new Pair(17,17);
		A[6] = new Pair(4,2);
		
		System.out.println(lis(A));
	}
	
	public static int lis(Pair[] A){
		Arrays.sort(A, new Comparator<Pair>() {
			@Override
			public int compare(Pair o1, Pair o2) {
				return o1.x - o2.x;
			}
		});
		
		int n = A.length;
		int max = 0;
		int[] dp = new int[n];
		Arrays.fill(dp, 1);
		
		for(int i=1; i<n; i++){
			for(int j=0; j<i; j++){
				if(A[i].y > A[j].y){
					dp[i] = Math.max(dp[i], dp[j]+1);
				}
			}
			max = Math.max(max, dp[i]);
		}
		
		return max;
	}
	
	public static class Pair{
		int x, y;
		public Pair(int x_, int y_){
			x = x_;
			y = y_;
		}
	}

}

http://www.geeksforgeeks.org/dynamic-programming-set-14-variations-of-lis/

http://people.csail.mit.edu/bdean/6.046/dp/