Ordered Subsequence

Time Limit: 4000/2000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)

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 解題思路：這道題和hdu 5542完全一樣，長度為m的遞增子序列的個數(shù)，樹狀數(shù)組+dp。 #include<iostream> #include<cstdio> #include<cstring> #include<set> #include<map> using namespace std;const int maxn = 10005; const int mod = 123456789; int n,m; int tree[105][maxn],a[maxn]; set<int> Set; map<int,int> Map;int lowbit(int x) {return x & -x; }void update(int i,int x,int c) {while(x <= n){tree[i][x] = (tree[i][x] + c) % mod;x += lowbit(x);} }int getsum(int i,int x) {int sum = 0;while(x > 0){sum = (sum + tree[i][x]) % mod;x -= lowbit(x);}return sum; }int main() {while(scanf("%d%d",&n,&m)!=EOF){Map.clear();Set.clear();memset(tree,0,sizeof(tree));for(int i = 1; i <= n; i++){scanf("%d",&a[i]);Set.insert(a[i]);}int cnt = 0;for(set<int>::iterator it = Set.begin(); it != Set.end(); it++)Map[*it] = ++cnt;for(int i = 1; i <= n; i++)for(int j = 1; j <= m; j++){int tmp = 0;if(j == 1) tmp = 1;else tmp = (tmp + getsum(j-1,Map[a[i]]-1)) % mod;update(j,Map[a[i]],tmp);}printf("%d\n",getsum(m,cnt));}return 0; }

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