Codeforces 1117G Recursive Queries 树状数组

Recursive Queries

刚开始把最大位置看成最大值， 我想这怎么写啊。。。

如果是最大位置的话， 我们能计算区间内每个值作为最大值的贡献，

然后拆成右边边第一个比它大的下标减去左边第一个比它大的下标， 然后离线之后就可以树状数组维护了。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 1e6 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

struct Bit {
    LL a[N];
    void init() {
        memset(a, 0, sizeof(a));
    }
    void modify(int x, int v) {
        if(!x) return;
        for(int i = x; i < N; i += i & -i) a[i] += v;
    }
    LL sum(int x) {
        LL ans = 0;
        for(int i = x; i; i -= i & -i) ans += a[i];
        return ans;
    }
    LL query(int L, int R) {
        return sum(R) - sum(L - 1);
    }
} bit[2];

int n, q, ql[N], qr[N];
int a[N], L[N], R[N];
int id[N];
LL ans[N];
stack<int> stk;
vector<int> qus[N];

bool cmpr(const int& a, const int& b) {
    return qr[a] < qr[b];
}
bool cmpl(const int& a, const int& b) {
    return ql[a] < ql[b];
}

int main() {
    scanf("%d%d", &n, &q);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    while(SZ(stk)) stk.pop();
    for(int i = 1; i <= n; i++) {
        while(SZ(stk) && a[stk.top()] < a[i]) stk.pop();
        L[i] = SZ(stk) ? stk.top() : 0;
        stk.push(i);
    }
    while(SZ(stk)) stk.pop();
    for(int i = n; i >= 1; i--) {
        while(SZ(stk) && a[stk.top()] < a[i]) stk.pop();
        R[i] = SZ(stk) ? stk.top() : n + 1;
        stk.push(i);
    }
    for(int i = 1; i <= q; i++) scanf("%d", &ql[i]), id[i] = i;
    for(int i = 1; i <= q; i++) scanf("%d", &qr[i]);
    sort(id + 1, id + 1 + q, cmpr);
    for(int i = 1, j = 1; i <= q; i++) {
        int x = id[i];
        while(j <= qr[x]) bit[0].modify(L[j], L[j]), bit[1].modify(L[j], 1), j++;
        LL val = bit[0].query(ql[x], qr[x]), cnt = bit[1].query(ql[x], qr[x]);
        ans[x] -= val + 1LL * (qr[x] - ql[x] + 1 - cnt) * (ql[x] - 1);
    }

    sort(id + 1, id + 1 + q, cmpl);
    bit[0].init(); bit[1].init();

    for(int i = q, j = n; i >= 1; i--) {
        int x = id[i];
        while(j >= ql[x]) bit[0].modify(R[j], R[j]), bit[1].modify(R[j], 1), j--;
        LL val = bit[0].query(ql[x], qr[x]), cnt = bit[1].query(ql[x], qr[x]);
        ans[x] += val + 1LL * (qr[x] - ql[x] + 1 - cnt) * (qr[x] + 1);
    }

    for(int i = 1; i <= q; i++) printf("%lld%c", ans[i] - (qr[i] - ql[i] + 1), " \n"[i == q]);
    return 0;
}

/*
*/