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

Note:

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;
}
};