首页 > 代码库 > hdu 4991 Ordered Subsequence(dp+树状数组)
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;
}
hdu 4991 Ordered Subsequence(dp+树状数组)
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。