DP_Wooden Sticks

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

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题意：求同时满足木棒的长宽都构成不下降子序列的最少组数。

思路：1、将木棒按照长（或宽）进行升序排序。
　　　2、建立递推数组dp[]，表示操作当前木棒时所需要的设定次数。
　　　3、dp[]的递推公式：　　若前面存在一个木棒的宽度大于当前木棒的宽度的时，当前木棒所需要进行的设置数一定会大于这个木棒所需要的设置数。
　　　　　　　　　　　当前木棒的dp[]=max{dp[前面宽度大于当前木棒宽度的木棒所对应的下标]}+1
　　　4、遍历数组dp[]，最大的dp即为题目所求最少需要的设定次数.
注意：排序完成后的木棒默认只设定一次（即假设排序完成后的序列为一个不下降序列）dp[]值初始化为1
　　　排序时在长度相同的情况下应优先按照宽度再次进行排序。

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#include<cstdio>
#include<algorithm>
using namespace std;
int dp[5005];
struct Node{
    int h,w;
}p[5005];
bool cmp( Node a, Node b){
    if( a.h!=b.h )
        return a.h<b.h ? true:false;
    return a.w<b.w ? true:false;
}
int main()
{
    int t,n;
    int maxn;

    scanf("%d",&t);
    while( t-- ){
        maxn=1;
        scanf("%d",&n);
        for( int i=0; i<n; i++)
            scanf("%d%d",&p[i].h,&p[i].w);
        sort( p, p+n, cmp);
        for( int j=0; j<n; j++)  
            dp[j]=1;
        for( int i=0; i<n; i++){
            for( int j=0; j<i; j++){
                if( p[j].w>p[i].w ){
                    dp[i]=max(dp[i],dp[j]+1);
                }
            }
        }
        for( int i=0; i<n; i++)
            if( dp[i]>maxn )
                maxn=dp[i];
        printf("%d\n",maxn);
    }

    return 0;
}

 

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感谢SLH的耐心讲解