动态规划中级教程 646. Maximum Length of Pair Chain

You are given n pairs of numbers. In every pair, the first number is always smaller than the second number.

Now, we define a pair (c, d) can follow another pair (a, b) if and only if b < c. Chain of pairs can be formed in this fashion. 

Given a set of pairs, find the length longest chain which can be formed. You needn't use up all the given pairs. You can select pairs in any order.

Example 1:

Input: [[1,2], [2,3], [3,4]]Output: 2Explanation: The longest chain is [1,2] -> [3,4]

Note:

1. The number of given pairs will be in the range [1, 1000].

说给我们一串链，找到最长的可以连接的链，条件是 【i】。frist 》 【j】。second

直接分析

【1，2】

dp【0】=1

【1，2】，【2，3】

dp【1】=1

【1，2】，【2，3】，【3，4】

dp【2】=dp【0】+1

写到这里大家可以发现这跟LIS基本是一样的这里已经可以写出状态转移方程

dp【i】=max（dp【i】，dp【j】+1）；

class Solution {public:    int findLongestChain(vector<vector<int>>& pairs) {        if(pairs.size()==0 )return 0;        sort(pairs.begin(),pairs.end());        int dp[pairs.size()];        for(int i=0;i<pairs.size();i++)        {            dp[i]=1;        }        int ans=1;        for(int i=1;i<pairs.size();i++)        {            for(int j=0;j<i;j++)            {                if(pairs[i][0]>pairs[j][1])                {                    dp[i]=max(dp[i],dp[j]+1);                    ans=max(ans,dp[i]);                }            }        }        return ans;    }};