Ordered Subsequence

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 114 Accepted Submission(s): 58

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

Source

BestCoder Round #8

官方题解

1003 Ordered Subsequence
首先数字有1万个，先离散化一下，把所有数字对应到1到n之间。这样对结果不影响。
dp[i][j]代表以第i个数字结尾上升子序列长度为j的种数。
dp[i][j]=sum{dp[k][j-1]}  for each a[k]<a[i]&&k<i
直接写循环会超时。需要优化。
可以用平衡树进行优化，上述的循环过程可以看成是一个区间求和过程。用线段树或者树状数组可以解决。
这样最终的复杂度是n*m*log(n)

这里我用100个数组数组搞了搞，注意中间的取模

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<map>
#include<ctime>
#define maxn 10010
#define LL long long
#define INF 999999
#define mod 123456789LL
using namespace std;struct node
{int id;LL val;bool operator <(const node&s)const{return val < s.val ;}
}qe[maxn];LL xt[101][maxn] ;
int n ,a[maxn] ;
LL dp[maxn][101] ;
void insert(int id,int x,int add)
{while( x <= n ){xt[id][x]+= add;if(xt[id][x] >= mod) xt[id][x] -= mod;x += (x&-x) ;}
}
LL sum(int id,int x)
{LL ans=0;while(x >0){ans += xt[id][x] ;if(ans>=mod) ans -= mod;x -= (x&-x) ;}return ans;
}
int main()
{int m,i,j ;while( scanf("%d%d",&n,&m) != EOF){for( i = 1 ; i <= n ;i++){scanf("%I64d",&qe[i].val) ;qe[i].id= i;}sort(qe+1,qe+1+n) ;j = 1 ;a[qe[1].id] = j ;for( i = 2 ; i <= n ;i++){if(qe[i].val==qe[i-1].val) a[qe[i].id]=j ;else a[qe[i].id] = ++j;}memset(xt,0,sizeof(xt)) ;memset(dp,0,sizeof(dp)) ;LL ans=0;for(i = 1 ; i <= n ;i++){dp[i][1] = 1 ;for( j = 2 ; j <= m && j <= i ;j++){dp[i][j] = sum(j-1,a[i]-1) ;}ans = (ans+dp[i][m])%mod;for( j = 1 ; j <= m && j <= i;j++){insert(j,a[i],dp[i][j]) ;}}cout << ans << endl;}return 0 ;
}

转载于:https://www.cnblogs.com/20120125llcai/p/3961621.html

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