ZOJ1025-Wooden Sticks（dp）

Wooden Sticks

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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. 


Sample Input 

3 
5 
4 9 5 2 2 1 3 5 1 4 
3 
2 2 1 1 2 2 
3 
1 3 2 2 3 1

Output for the Sample Input

2 
1 
3

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Source: Asia 2001, Taejon (South Korea)

思路：显然需要将数据处理一下，将其按木棍长度升序排列，若相同则按重量升序排列；

则本题转化为求最大上升子序列的最小个数，显然这又能转化为求最大下降子序列的长度；

ps:最大下降子序列的中每一个元素，都必须单独在一个不下降子序列里，这样最大下降子序列的长度就是最大不下降子序列的最少个数

代码：

#include <bits/stdc++.h>using namespace std;const int maxN=5001;int i,k,test,n;struct stick { int l; int w; }data[maxN]; void solve(){    int dp[maxN];    int res=0;      for(int i=0;i<n;i++)      {          dp[i]=1;          for(int j=0;j<i;j++)              if(data[j].w>data[i].w)                  dp[i]=max(dp[i],dp[j]+1);          res=max(res,dp[i]);//更新      }      printf("%d\n",res);  }  bool cmp(stick a,stick b) { if(a.l==b.l)    return a.w<b.w;return a.l<b.l;} int main() { scanf("%d", &test); while(test--){ scanf("%d", &n); for(i = 0; i < n; ++i) scanf("%d%d", &data[i].l, &data[i].w); sort(data,data+n,cmp); solve();} return 0; }