HDU-1051 一个DP问题

Problem Description

There is a pile of n wooden sticks. The length and weight of each stick are known in advance. The sticks are to be processed by a woodworking machine in one by one fashion. It needs some time, called setup time, for the machine to prepare processing a stick. The setup times are associated with cleaning operations and changing tools and shapes in the machine. The setup times of the woodworking machine are given as follows:

(a) The setup time for the first wooden stick is 1 minute.
(b)
Right after processing a stick of length l and weight w , the machine
will need no setup time for a stick of length l' and weight w' if
l<=l' and w<=w'. Otherwise, it will need 1 minute for setup.

You
are to find the minimum setup time to process a given pile of n wooden
sticks. For example, if you have five sticks whose pairs of length and
weight are (4,9), (5,2), (2,1), (3,5), and (1,4), then the minimum setup
time should be 2 minutes since there is a sequence of pairs (1,4),
(3,5), (4,9), (2,1), (5,2).

Input

The
input consists of T test cases. The number of test cases (T) is given
in the first line of the input file. Each test case consists of two
lines: The first line has an integer n , 1<=n<=5000, that
represents the number of wooden sticks in the test case, and the second
line contains n 2 positive integers l1, w1, l2, w2, ..., ln, wn, each of
magnitude at most 10000 , where li and wi are the length and weight of
the i th wooden stick, respectively. The 2n integers are delimited by
one or more spaces.

Output

The output should contain the minimum setup time in minutes, one per line.

分析

　　这道题是问将一组木条划分一下，要求每个链中长度和重量是不下降的，求最少能分出几个链，好像挺眼熟的感觉？没错这不是导弹拦截吗？如果你对迪尔沃斯（音译）定理了解足够深的话，接下来怎么做就很明了了，最少链的划分=最长反链的长度，数据n<=5000，n2就可以不需要加优化。

　　如果你很好奇定理证明，自行百度……

　　

 #include<iostream>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 const int N=1e5+;
 int dp[N];
 struct Node{
     int l,w;
     bool operator <(const Node &A)const{
         if(l==A.l){
             if(w==A.w)return ;
             return w<A.w;
         }
         return l<A.l;
     }
 }a[N];
 int main(){
     int t;
     cin>>t;
     while(t--){
         int n;
         memset(dp,,sizeof(dp));
         cin>>n;
         for(int i=;i<=n;i++){
             cin>>a[i].l>>a[i].w;
         }
         sort(a+,a+n+);
         int ans=;
         for(int i=;i<=n;i++){
             dp[i]=;
             for(int j=;j<i;j++)
                 if(a[j].w>a[i].w)
                     dp[i]=max(dp[j]+,dp[i]);
             ans=max(dp[i],ans);
         }
         cout<<ans<<endl;
     }
 }