POJ 1631 Bridging signals

事实上就是求最长上升子序列的长度。用nlogn的写法。我用了线段树。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

const int maxn=40000+10;
int dp[maxn];
int a[maxn];
int T,n,ans,f;
struct SegTree
{
    int MAX;
}s[4*maxn];

void pushUp(int rt)
{
    s[rt].MAX=max(s[2*rt].MAX,s[2*rt+1].MAX);
}

void build(int l,int r,int rt)
{
    s[rt].MAX=0;
    if(l==r) return;
    int m=(l+r)/2;
    build(l,m,2*rt);
    build(m+1,r,2*rt+1);
}

void update(int pos,int val,int l,int r,int rt)
{
    if(l==r)
    {
        s[rt].MAX=val;
        return;
    }

    int m=(l+r)/2;
    if(pos<=m) update(pos,val,l,m,2*rt);
    else update(pos,val,m+1,r,2*rt+1);
    pushUp(rt);
}

void quary(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        f=max(f,s[rt].MAX);
        return;
    }

    int m=(l+r)/2;
    if(L<=m) quary(L,R,l,m,2*rt);
    if(R>m) quary(L,R,m+1,r,2*rt+1);
}

int main()
{
 //   freopen("test.in","r",stdin);
 //   freopen("test.out","w",stdout);

    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n); ans=0;
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        build(1,n,1);
        for(int i=1;i<=n;i++)
        {
            f=0;
            if(a[i]==1) f=0;
            else quary(1,a[i]-1,1,n,1);
            dp[i]=f+1;
            update(a[i],dp[i],1,n,1);
            ans=max(ans,dp[i]);
        }
        printf("%d\n",ans);
    }
    return 0;
}