Problem Description
I konw you guys have solve so many problems about increasing sequence, this time, a little change has been made.
Assume that there is a sequence S = {s1, s2, s3, ..., sn}, si = (xi, yi).You should find two increasing subsequence L1 and L2, and they have no common elements, means L1∩L2 = φ, and the sum of their lenth is as max as possible.
Here we assume si > sj is that (xi > xj && yi > yj) or (xi >= xj && yi > yj) or (xi > xj && yi >= yj). I will ensure that all elements' coordinates are distinct, i.e., si != sj (i!=j).
Input
The input consists of multiple test cases. Each case begins with a line containing a positive integer n that is the length of the sequence S, the next n lines each contains a pair integers (xi, yi), i = 1,...n.1 <= n <= 5000,1<=xi,yi<=2^31.
Output
For each test case, output one line containing the the maximum sum of the two increasing subsequence L1 and L2 you can find.
Sample Input
3
1 3
3 1
2 2
4
1 2
2 1
4 3
3 4
Sample Output
2
4
问题描述
我知道你们已经解决了许多关于增加序列的问题，这一次，已经做了一点改变。
假设有一个序列S = {s1，s2，s3，...，sn}，si =（xi，yi）。你应该找到两个增加的子序列L1和L2，它们没有共同的元素，意味着L1∩ L2 =φ，并且它们的长度之和尽可能地最大。
这里我们假设si> sj是（xi> xj && yi> yj）或（xi> = xj && yi> yj）或（xi> xj && yi> = yj）。 我将确保所有元素的坐标是不同的，即si！= sj（i！= j）。
输入
输入包含多个测试用例。 每个案例都以包含正整数n的行开始，该正整数n是序列S的长度，接下来的n行每行包含一对整数（xi，yi），i = 1，... n.1 <= n <=5000,1<= XI，易<=2 ^ 31。
输出
对于每个测试用例，输出一行，其中包含您可以找到的两个递增子序列L1和L2的最大总和。
Sample Input
3
1 3
3 1
2 2
4
1 2
2 1
4 3
3 4
Sample Output
2
4
为了回答问题，先把您的问题机翻了一遍，不是很懂您想问什么。但关于子集序列的最大连续值问题的算法，我还是有一些想法的。
给您举个栗子：
如给定序列{ -2, 11, -4, 13, -5, -2 }，其最大连续子序列为{11,-4,13}，最大连续子序列和即为20。
那么我们可以直接入手代码，求出所有可能连续子列的和即可：
int MaxSubSequm1(int A[], int N)
{
int oneSum, maxSum = 0;
int i, j, k;
for (i = 0; i < N; i++) {
for (j = i; j < N; j++){
oneSum = 0;
for (k = i; k <= j; k++)
oneSum += A[k];
if (oneSum>maxSum)
maxSum = oneSum;
}
}
return maxSum;
}
我只能做到一种方法，这边有一位博主的详尽解答，希望对您有所帮助。
https://blog.csdn.net/zxc123e/article/details/79679488