poj1065Wooden Sticks(dp——最长递减数列)

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 ( 9 , 4 ) , ( 2 , 5 ) , ( 1 , 2 ) , ( 5 , 3 ) , and ( 4 , 1 ) , then the minimum setup time should be 2 minutes since there is a sequence of pairs ( 4 , 1 ) , ( 5 , 3 ) , ( 9 , 4 ) , ( 1 , 2 ) , ( 2 , 5 ) .

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 2n 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 

Sample Output

2
1
3

先附上AC代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
pair<int ,int > a[5000];

int cmp(pair<int ,int >a,pair<int ,int > b ) {
if(a.first==b.first) return a.second<b.second;
else return a.first<b.first;
}
int main(){
int T,n;
cin>>T;
int dp[5005];
while(T--){
memset(dp,0,sizeof(dp));
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d%d",&a[i].first,&a[i].second);
sort(a,a+n,cmp); //对第一属性进行排序，然后第二属性求最长递减数列，长度即为所求（与之前做过的最少拦截系统类似）
for(int i=0;i<n;i++){
for(int j=0;j<i;j++){
if(a[i].second<a[j].second) dp[i]=max(dp[i],dp[j]+1);
}
if(dp[i]==0) dp[i]=1;
}
int maxn=0;
for(int i=0;i<n;i++)
if(dp[i]>maxn) maxn=dp[i];
printf("%d ",maxn);
}
return 0;
}

这道题想了很久都没有思路，最后发现与最少拦截系统类似，不过就是相当于一个改编而已，而且是属于那种换汤不换药的改编。

在做ACM题时，我觉得应该多在头脑中积累一些典型的例题，毕竟即便是出题人也不可能凭空就出来一道题（这种可能非常非常小），出题人肯定也是根据现有的题进行加工改变，而如果我们积累了一些典型的例题，未必不能很快的解出答案，实现AC.