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hdu 4991（dp+树状数组）

2024-07-21 22:31:34   214人阅读

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=4991

Ordered Subsequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 221    Accepted Submission(s): 112


Problem Description
A numeric sequence of ai is ordered if a1<a2<……<aN. Let the subsequence of the given numeric sequence (a1, a2,……, aN) be any sequence (ai1, ai2,……, aiK), where 1<=i1<i2 <……<iK<=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
 
第一遍做用的是爆搜，代码很短，果断超时，然后就愣了，后来才知道是用树状数组，我用树状数组用的比较少，想了好久终于想懂了；

思路： dp[i][j]表示一第i个数结尾，长度为j的序列的个数；

那么状态转移方程式：dp[i][j]=sum(dp[k][j-1]);(k<i&&a[i]>a[k])

如果直接遍历，那么时间复杂度至少是O（n*n），超时，所以考虑用树状数组维护一段区间的和；

#include <iostream>
#include <stdio.h>
#include <string>
#include <string.h>
#include <cstdio>
#include <cmath>
#include <algorithm>
typedef long long ll;
const int mod=123456789;
using namespace std;
ll a[10100],b[10100],c[10100],dp[10100][110];
int  n,m;
int lowbit(int t)
{
 return t&(-t);
}
void modify(int t,ll d)
{
  while(t<=n)
  {
   c[t]+=d;
   t+=lowbit(t);
  }
}
ll getsum(int t)
{
  ll ans=0;
  while(t>0)
  {
    ans=(ans+c[t])%mod;
    t-=lowbit(t);
  }
  return ans;
}

int main()
{
        while(scanf("%d%d",&n,&m)!=EOF)
        {
          for(int i=1;i<=n;i++)
          {
            scanf("%I64d",&a[i]);
            b[i]=a[i];
          }
          sort(b+1,b+1+n);
          memset(dp,0,sizeof(dp));
          for(int i=1;i<=n;i++)dp[i][1]=1;
          for(int j=2;j<=m;j++)
          {
            memset(c,0,sizeof(c));
            for(int i=1;i<=n;i++)
            {
             int indx=lower_bound(b+1,b+1+n,a[i])-b;
             dp[i][j]=getsum(indx-1);
             modify(indx,dp[i][j-1]);
            }
          }
          ll cnt=0;
          for(int i=1;i<=n;i++)
          {
           cnt=(cnt+dp[i][m])%mod;
          }
          printf("%I64d\n",cnt%mod);
        }

        return 0;
}


hdu 4991（dp+树状数组）


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