POJ1065Wooden Sticks[DP LIS]

Wooden Sticks

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 21902		Accepted: 9353

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

Source

Taejon 2001

--------------------------

和poj3636同样的道理

因为偏序关系是<=，所以w从小到大相同l小的在前，找最长下降子序列

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int N=,INF=1e9;
struct data{
    int w,l;
}da[N];
bool cmpda(data a,data b){
    if(a.w>b.w) return ;
    if(a.w<b.w) return ;
    if(a.w==b.w) return a.l<b.l?:;
    return ;
}
int t,n;
int f[N],g[N],a[N];
bool cmp(int a,int b){
    return a>b;
}
int dp(){
    int ans=;
    sort(da+,da++n,cmpda);
    memset(f,,sizeof(f));
    for(int i=;i<=n;i++) g[i]=-INF,a[i]=da[i].l;
    for(int i=;i<=n;i++){
        int k=lower_bound(g+,g++n,a[i],cmp)-g;
        f[i]=k;
        g[k]=a[i];
        ans=max(ans,f[i]);
    }
    return ans;
}

int main(int argc, const char * argv[]) {
    scanf("%d",&t);
    for(int i=;i<=t;i++){
        scanf("%d",&n);
        for(int i=;i<=n;i++) scanf("%d%d",&da[i].l,&da[i].w);
        printf("%d\n",dp());
    }
    return ;
}