hdu 4991 Ordered Subsequence(dp+树状数组)

题目链接：hdu 4991 Ordered Subsequence

题目大意：给定一个序列，求有多少个子序列满足长度为m，并且递增。

解题思路：dp[i][j]表示说选了以第i个数为结尾，长度为j的递增子串方案数。将每个数离散化后用树状数组维护即可。

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

#define lowbit(x) ((x)&(-x))

using namespace std;
typedef long long ll;
const int maxn = 10005;
const int maxm = 105;
const ll mod = 123456789;

struct state {
    ll val;
    int pos, rank;
    state (ll val = 0, int pos = 0, int rank = 0) {
        this->val = val;
        this->pos = pos;
        this->rank = rank;
    }
};

int N, M, T;
ll dp[maxn][maxm], fenw[maxm][maxn];
vector<state> vec;

inline bool sort_val(const state& a, const state& b) {
    return a.val < b.val;
}

inline bool sort_pos(const state& a, const state& b) {
    return a.pos < b.pos;
}

void add (ll* f, int x, ll v) {
    while (x <= T) {
        f[x] = (f[x] + v) % mod;
        x += lowbit(x);
    }
}

ll sum (ll* f, int x) {
    ll ret = 0;
    while (x) {
        ret = (ret + f[x]) % mod;
        x -= lowbit(x);
    }
    return ret;
}

void init () {

    ll x;
    vec.clear();
    for (int i = 1; i <= N; i++) {
        scanf("%lld", &x);
        vec.push_back(state(x, i));
    }

    sort(vec.begin(), vec.end(), sort_val);

    vec[0].rank = 2;
    for (int i = 1; i < N; i++) {
        vec[i].rank = vec[i-1].rank;
        if (vec[i].val != vec[i-1].val)
            vec[i].rank++;
    }
    T = vec[N-1].rank;

    sort(vec.begin(), vec.end(), sort_pos);
}

ll solve () {
    memset(dp, 0, sizeof(dp));
    memset(fenw, 0, sizeof(fenw));

    add(fenw[0], 1, 1);

    for (int i = 1; i <= N; i++) {
        for (int j = min(i, M); j >= 1; j--) {
            ll tmp = sum(fenw[j-1], vec[i-1].rank - 1);
            dp[i][j] = (dp[i][j] + tmp) % mod;
            add(fenw[j], vec[i-1].rank, tmp);
        }
    }

    ll ret = 0;
    for (int i = 1; i <= N; i++)
        ret = (ret + dp[i][M]) % mod;
    return ret;
}

int main () {
    while (scanf("%d%d", &N, &M) == 2) {
        init();
        printf("%lld\n", solve());
    }
    return 0;
}