树状数组优化+dp

http://acm.hdu.edu.cn/showproblem.php?pid=5542

题意：给出长度为N(1 <= N <= 1e3)的序列，求所有长度为M的递增子序列的个数。

解法：容易想到dp[i][j]以表示第i个元素结尾，lis为j的数量，状态转移方程：dp[i][k] = ∑dp[j][k-1] 1<=j< i, a[j] < a[i]. 时间复杂度为O（n³）会超时。

树状数组优化。构造dp[a[i]][j]以a[i]这个数结尾的lis长度为j的数量。因为a[i]数据范围1e9太大需要离散化。树状数组维护dp[a[i]][j] = dp[a[i]-1][j-1] + ..... dp[1][j-1]。1到a[i]-1 lis长度为j-1的数量

https://www.cnblogs.com/dilthey/p/9898230.html

//#include<bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
#include <stdlib.h>
using namespace std;
typedef long long ll ;
#define int ll
#define mod 1000000007
#define gcd(m,n) __gcd(m, n)
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j)  for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
//int lcm(int a , int b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define INF  0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define pb push_back
#define mp make_pair
#define all(v) v.begin(),v.end()
#define size(v) (int)(v.size())
#define cin(x) scanf("%lld" , &x);
const int N = 1e6+9;
const int maxn = 1e3+5;
const double esp = 1e-6;
int a[maxn] , b[maxn] , len;
int dp[maxn][maxn];
int cnt ;
int n , m ;
int lowerbit(int x){
    return x&(-x);
}
void add(int x , int y , int val){
    while(x <= len){
        dp[x][y] += val;
        dp[x][y] %= mod;
        x += lowerbit(x);
    }
}
int getsum(int x , int y){
    int ans = 0 ;
    while(x){
        ans += dp[x][y];
        ans %= mod ;
        x -= lowerbit(x);
    }
    return ans ;
}

void solve(){
    ME(dp , 0);

    scanf("%lld%lld" , &n , &m);
    rep(i , 1 , n){
        scanf("%lld" , &a[i]);
        b[i] = a[i];
    }
    sort(b+1 , b+1+n);
    len = unique(b+1 , b+1+n) - b - 1 ;
    rep(i , 1 , n){
        rep(j , 1 , m){
            int pos = lower_bound(b+1 , b+1+len , a[i]) - b ;
            if(j == 1){
                add(pos , j , 1);
            }else{
                add(pos , j , getsum(pos-1 , j-1));
            }
        }
    }
    cout << "Case #" << ++cnt << ": " << getsum(len , m) << endl;
}

signed main()
{
    //ios::sync_with_stdio(false);
    int t ;
    cin >> t ;
    while(t--){
        solve();
    }
}