hdu 4991 Ordered Subsequence

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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 DescriptionA 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. 

 

InputMulti 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 

 

OutputFor each case, output answer % 123456789. 

 

Sample Input3 21 1 27 31 7 3 5 9 4 8 

 

Sample Output212 

 

SourceBestCoder Round #8官方題解

1003 Ordered Subsequence首先數字有1萬個，先離散化一下，把所有數字對應到1到n之間。這樣對結果不影響。dp[i][j]代表以第i個數字結尾上升子序列長度為j的種數。dp[i][j]=sum{dp[k][j-1]}  for each a[k]<a[i]&&k<i直接寫迴圈會逾時。需要最佳化。可以用平衡樹進行最佳化，上述的迴圈過程可以看成是一個區間求和過程。用線段樹或者樹狀數組可以解決。這樣最終的複雜度是n*m*log(n)

這裡我用100個數組數組搞了搞，注意中間的模數

#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 123456789LLusing 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 ;}

　　

hdu 4991 Ordered Subsequence