POJ - 1065 Wooden Sticks(DP 子序列问题）

Wooden Sticks

Time Limit: 1000MS Memory Limit: 10000KB 64bit IO Format: %lld & %llu

Submit Status

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

213

这题可以用贪心做也可以用DP做

DP：最少上升子序列数等于最长下降子序列长度，，注意过程中保留最大值，，

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int N = 50100;
int dp[N];
struct node
{
    int x, y;
}p[N];
int cmp(node a,node b)
{
    if(a.x!=b.x)
        return a.x<b.x;
    else
        return a.y<b.y;
}


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