1748E Yet Another Array Counting Problem

1748E Yet Another Array Counting Problem

题目传送门

题目大意：

现在有一个长度为$n$的数组$a$ , 现在你要找到一个数组$b$的每一个数都$<=m$ 使得在$n$中的任意区间$[l,r]$满足$ma{x}_{i=l}^r{a}_i=ma{x}_{i=l}^r{b}_i$ 求方案数 $ mod 1e9 + 7$

做法

设$d{p}_{pos,i}$表示$pos$作为左边最大值的位置，且$b_{pos}<=i$ 的方案数
显然$d{p}_{pos,i}={\sum }_{j=1}^{i}d{p}_{lpos,j-1}\ast d{p}_{rpos,j}$
可以用线段树维护$lpos、rpos$

code

#include
#define fu(x , y , z) for(int x = y ; x <= z ; x ++) 
#define LL long long 
using namespace std;
const int N = 2e5 + 5;
const LL mod = 1e9 + 7;
vector<LL> dp[N];
int a[N] , n , m , rt , tr[N << 2];
int read () {
    int val = 0 , fu = 1;
    char ch = getchar ();
    while (ch < '0' || ch > '9') {
        if (ch == '-') fu = -1;
        ch = getchar ();
    }
    while (ch >= '0' && ch <= '9') {
        val = val * 10 + (ch - '0');
        ch = getchar ();
    }
    return val * fu;
}
void build (int p , int l , int r) {
    if (l == r) {
        tr[p] = l;
    }
    else {
        int mid = l + r >> 1 , ll = p << 1 , rr = (p << 1) + 1;
        build (ll , l , mid) , build (rr , mid + 1 , r);
        if (a[tr[ll]] >= a[tr[rr]]) {
            tr[p] = tr[ll];
        }
        else 
            tr[p] = tr[rr];
    }
}
int query (int p , int l , int r , int L , int R) {
    if (L <= l && R >= r) {
        return tr[p];
    }
    else {
        int mid = l + r >> 1 , ans1 = 0 , ans2 = 0 , ll = p << 1 , rr = (p << 1) + 1;
        if (L <= mid) 
            ans1 = query (ll , l , mid , L , R);
        if (mid < R)
            ans2 = query (rr , mid + 1 , r , L , R);
        return a[ans1] >= a[ans2] ? ans1 : ans2;
    }
}
int dfs (int l , int r) {
    if (l > r) 
        return -1;
    if (l == r) {
        fu(i , 1 , m) 
            dp[l][i] = i;
        return l;
    }
    int id = query (1 , 1 , n , l , r);
    int ll = dfs (l , id - 1) , rr = dfs (id + 1 , r);
    if (ll == -1)
        fu(i , 1 , m)
            dp[id][i] = (dp[rr][i] + dp[id][i - 1]) % mod;
    else if (rr == -1) 
        fu(i , 1 , m) 
            dp[id][i] = (dp[ll][i - 1] + dp[id][i - 1]) %mod;
    else 
        fu(i , 1 , m) 
            dp[id][i] = (dp[ll][i - 1] * dp[rr][i] % mod + dp[id][i - 1]) % mod;
    return id;
}
int main () {
    int T = read ();
    while (T --) {
        n = read () , m = read ();
        fu(i , 1 , n) {
            a[i] = read ();
            dp[i].clear ();
            dp[i].resize(m + 1);
        }
        build (1 , 1 , n);
        rt = dfs (1 , n);
        printf ("%lld\n" , dp[rt][m]);
    }
    return 0;
}