hdu 4991(树状数组+DP)

Ordered Subsequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 464    Accepted Submission(s): 216

Problem Description

A numeric sequence of a_i is ordered if a₁<a₂<……<a_N. Let the subsequence of the given numeric sequence (a₁, a₂,……, a_N) be any sequence (a_i1, a_i2,……, a_iK), where 1<=i₁<i₂ <……<i_K<=N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, eg. (1, 7), (3, 4, 8) and many others.

Your program, when given the numeric sequence, must find the number of its ordered subsequence with exact m numbers.

 

 

Input

Multi test cases. Each case contain two lines. The first line contains two integers n and m, n is the length of the sequence and m represent the size of the subsequence you need to find. The second line contains the elements of sequence - n integers in the range from 0 to 987654321 each.
Process to the end of file.
[Technical Specification]
1<=n<=10000
1<=m<=100

 

 

Output

For each case, output answer % 123456789.

 

 

Sample Input

3 2 1 1 2 7 3 1 7 3 5 9 4 8

 

 

Sample Output

2 12

 

题意:问在 长度为n的串中有多少个长度为 m 的上升子序列.

题解:dp[i][j]代表以第 i 个元素结尾，长度为 j 的子序列 的个数。那么 dp[i][j] = sum(dp[k][j-1]) (1<=k<j).

但是这样的话我们直接枚举是 O（n*n*m）这样的时间复杂度是接受不了的。所以求和的那一层用树状数组优化成 O(n*log(n)*m)

/**
状态转移方程:
dp[i][j] = sum(dp[k][j-1]) (1<=k<i&&a[k]<a[i])
*/
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
typedef long long LL;
const LL mod = 123456789;
const int N = 10005;
const int M = 105;
int cnt,n,m;
LL dp[N][M]; ///dp[i][j]代表第 i 个元素结尾,长度为 j 的递增子序列个数.
LL c[N],b[N],a[N];

LL lowbit(int i){
    return i&(-i);
}
void update(int idx,int x,LL v){
    for(int i=idx;i<=cnt;i+=lowbit(i)){
        dp[i][x]=(dp[i][x]+v)%mod;
    }
}
LL getsum(int idx,int x){
    LL sum = 0;
    for(int i=idx;i>0;i-=lowbit(i)){
        sum=(sum+dp[i][x])%mod;
    }
    return sum;
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        for(int i=1;i<=n;i++){
            scanf("%lld",&a[i]);
            b[i] = a[i];
        }
        cnt = 1;
        for(int i=2;i<=n;i++){ ///离散化 a 数组对应树状数组的 1 - cnt
            if(b[i]!=b[i-1]){
                b[++cnt] = b[i];
            }
        }
        sort(b+1,b+cnt+1);
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++){
            int idx = lower_bound(b+1,b+1+cnt,a[i])-b;
            for(int j=1;j<=m;j++){
                LL v;
                if(j == 1) v = 1;
                else v = getsum(idx-1,j-1);
                update(idx,j,v);
            }
        }
        LL ans = getsum(cnt,m);
        printf("%lld\n",ans);
    }
    return 0;
}

 

    #include<cstdio>
    #include<string.h>
    #include<algorithm>
    using namespace std;
    #define LL long long
    const int mod=123456789;
    int n,m;
    int a[10005];
    int b[10005],cnt;

    inline void Add(int &a,int b){
        a=(a+b)%mod;
    }
    inline int lowbit(int x){
        return x&(-x);
    }
    int sum[101][10005];
    inline void add(int id,int x,int v){
        while(x<=cnt){
            Add(sum[id][x],v);
            x+=lowbit(x);
        }
    }
    inline int query(int id,int x){
        int ans=0;
        while(x){
            Add(ans,sum[id][x]);
            x-=lowbit(x);
        }
        return ans;
    }

    int main(){
        while(~scanf("%d%d",&n,&m)){
            cnt=0;
            for(int i=1;i<=n;i++){
                scanf("%d",&a[i]);
                b[++cnt]=a[i];
            }
            sort(b+1,b+cnt+1);
            cnt=(int)(unique(b+1,b+1+cnt)-(b+1));
            for(int i=1;i<=n;i++)a[i]=(int)(lower_bound(b+1,b+1+cnt,a[i])-b);

            memset(sum,0,sizeof sum);

            int ans=0;
            for(int i=1;i<=n;i++){
                for(int j=0;j<m;j++){
                    if(a[i]>1){
                        int sum=query(j,a[i]-1);
                        add(j+1,a[i],sum);
                    }
                }
                add(1,a[i],1);
            }
            ans=query(m,cnt);
            printf("%d\n",ans);
        }
        return 0;
    }