POJ3636Nested Dolls[DP LIS]

Nested Dolls

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8323		Accepted: 2262

Description

Dilworth is the world's most prominent collector of Russian nested dolls: he literally has thousands of them! You know, the wooden hollow dolls of different sizes of which the smallest doll is contained in the second smallest, and this doll is in turn contained in the next one and so forth. One day he wonders if there is another way of nesting them so he will end up with fewer nested dolls? After all, that would make his collection even more magnificent! He unpacks each nested doll and measures the width and height of each contained doll. A doll with width w₁ and height h₁ will fit in another doll of width w₂ and height h= if and only if w₁ < w₂ and h₁ < h₂. Can you help him calculate the smallest number of nested dolls possible to assemble from his massive list of measurements?

Input

On the first line of input is a single positive integer 1 ≤ t ≤ 20 specifying the number of test cases to follow. Each test case begins with a positive integer 1 ≤ m ≤ 20000 on a line of itself telling the number of dolls in the test case. Next follow 2m positive integers w₁, h₁,w₂, h₂, ... ,w_m, h_m, where w_i is the width and h_i is the height of doll number i. 1 ≤ w_i, h_i ≤ 10000 for all i.

Output

For each test case there should be one line of output containing the minimum number of nested dolls possible.

Sample Input

4
3
20 30 40 50 30 40
4
20 30 10 10 30 20 40 50
3
10 30 20 20 30 10
4
10 10 20 30 40 50 39 51

Sample Output

1
2
3
2

Source

Nordic 2007

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貌似有一种很厉害的东西叫Dilworth定理

最少的chain个数等于最大的antichain的大小

按照w从小到大排序，剩下考虑h，chain应该是严格单增

最少几个，antichain，不就是导弹拦截的第二问，最长不下降子序列

要注意的是w相同时h大的放在前，因为w相同嵌套是违法的，WA好几次

 

//
//  main.cpp
//  poj3636
//
//  Created by abc on 16/8/30.
//  Copyright © 2016年 abc. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int N=20005,INF=1e9;
struct data{
    int w,h;
}da[N];
bool cmpda(data a,data b){
    if(a.w>b.w) return 0;
    if(a.w<b.w) return 1;
    if(a.w==b.w) return a.h>b.h?1:0;
    return 1;
}
int t,n;
int f[N],g[N],a[N];
bool cmp(int a,int b){
    return a>b;
}
int dp(){
    int ans=0;
    sort(da+1,da+1+n,cmpda);
    memset(f,0,sizeof(f));
    for(int i=1;i<=n;i++) g[i]=-INF,a[i]=da[i].h;
    for(int i=1;i<=n;i++){
        int k=upper_bound(g+1,g+1+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=1;i<=t;i++){
        scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%d%d",&da[i].w,&da[i].h);
        printf("%d
",dp());
    }
    return 0;
}