hdu magic balls

magic balls

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 74    Accepted Submission(s): 10

Problem Description

The town of W has N people. Each person takes two magic balls A and B every day. Each ball has the volume ai and bi . People often stand together. The wizard will find the longest increasing subsequence in the ball A. The wizard has M energy. Each point of energy can change the two balls’ volume.(swap(ai,bi) ).The wizard wants to know how to make the longest increasing subsequence and the energy is not negative in last. In order to simplify the problem, you only need to output how long the longest increasing subsequence is.

 

 

Input

The first line contains a single integer T(1≤T≤20) (the data for N>100 less than 6 cases), indicating the number of test cases.
Each test case begins with two integer N(1≤N≤1000) and M(0≤M≤1000) ,indicating the number of people and the number of the wizard’s energy. Next N lines contains two integer ai and bi(1≤ai,bi≤109) ,indicating the balls’ volume.

 

 

Output

For each case, output an integer means how long the longest increasing subsequence is.

 

 

Sample Input

2 5 3 5 1 4 2 3 1 2 4 3 1 5 4 5 1 4 2 3 1 2 4 3 1

 

 

Sample Output

4 4

 

 

Source

BestCoder Round #20 

 

思路： 建 m 棵线段树，里面维护的是权值，

PS :BC的pst好像都好水，写的时候那么大错误还过了。。

==

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <algorithm>
#define LL long long
#define maxn 2010
#define INF 0x3f3f3f3f
#define mod 1000000007
using namespace std;

int a[maxn],b[maxn],id[maxn] ;
int  ql,qr;
short v ;
struct node
{
    int n,mid,hehe[maxn];
    short Max[maxn*4];
    void init(int n )
    {
       for(int i=1;i<=n*4;i++)
        Max[i]=0;
       for(int i=1;i<=n;i++)
        hehe[i]=0;
    }
    void insert(int L,int R,int o )
    {
        if(L==R)
        {
            Max[o]=v;
            hehe[L]=v;
            return ;
        }
        mid=(L+R)>>1 ;
        if(ql<=mid) insert(L,mid,o<<1) ;
        else insert(mid+1,R,o<<1|1) ;
        Max[o]=max(Max[o<<1],Max[o<<1|1]);
    }
    void find(int L,int R,int o)
    {
        if(ql<=L&&qr>=R)
        {
            v=max(Max[o],v) ;
            return ;
        }
        mid=(L+R)>>1 ;
        if(ql<=mid) find(L,mid,o<<1) ;
        if(qr>mid) find(mid+1,R,o<<1|1) ;
    }
}qe[1010] ;
int main()
{
    int i,xx,yy,ans ;
    int j,n,m,pre,sz;
    int T,tmp,val;
    //freopen("in.txt","r",stdin);
    cin >> T ;
    while(T--)
    {
        scanf("%d%d",&n,&m) ;
        sz=0;
        for( i = 1 ; i <= n ;i++){
            scanf("%d%d",&a[i],&b[i]) ;
            id[sz++]=a[i];
            id[sz++]=b[i];
        }
        sort(id,id+sz) ;
        sz=unique(id,id+sz)-id;
        ans=0;
        for( i = 0; i <= m;i++)
        {
            qe[i].init(sz);
        }
        for( i = 1 ; i <= n ;i++)
        {
            a[i]=lower_bound(id,id+sz,a[i])-id;
            b[i]=lower_bound(id,id+sz,b[i])-id;
            a[i]++;b[i]++;
            for( j = min(i,m); j >= 0 ;j--)
            {
                ql=1;
                qr=a[i]-1;
                v=0;
                if(ql<=qr) qe[j].find(1,sz,1) ;
                ql=a[i] ;
                v=v+1;
                tmp=v;
                if(qe[j].hehe[ql]<v)qe[j].insert(1,sz,1);
                if(j>0){
                    ql=1;
                    qr=b[i]-1;
                    v=0;
                    if(ql<=qr) qe[j-1].find(1,sz,1) ;
                    ql=b[i] ;
                    v=v+1;
                    tmp=max(tmp,(int)v);
                    if(qe[j].hehe[ql]<v)qe[j].insert(1,sz,1);
                }
                ans=max(ans,tmp);
            }
        }
        cout<<ans<<endl;
    }
    return 0 ;
}