[2019杭电多校第十场][hdu6701]Make Rounddog Happy

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6701

题目大意为求满足 $max(a_{l},a_{l+1}cdot cdot cdot a_{r})-(r-l+1)<=k$的区间个数。

先预处理出前缀最大值和后缀最大值和ST表，然后分治。

每次可以得到这次分治区间的区间最大值，然后我们要求出以该最大值为区间最大值时的合法区间数目。

这里我们可以枚举合法区间的左端点(右端点)，然后通过化简上式得到合法的右端点(左端点)，选择枚举左还是右端点由当前区间最大值的位置所决定，因为左端点一定在最大值左边，右端点一定在最大值右边，所以选择数量较小的一边来枚举。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 3e5 + 110;
int a[maxn], pos[maxn], L[maxn], R[maxn], Log[maxn];
int n, k;
int dp[maxn][20];
int query(int l, int r) {
    int k = Log[r - l + 1];
    if (a[dp[l][k]] > a[dp[r - (1 << k) + 1][k]])
        return dp[l][k];
    else
        return dp[r - (1 << k) + 1][k];
}
ll dfs(int l, int r) {
    if (l == r)return a[l] - 1 <= k;
    if (l > r)return 0;
    int cnt = query(l, r);
    ll ans = 0;
    if (cnt - l < r - cnt) {
        for (int i = l; i <= cnt; i++) {
            int ls = max(cnt, i + a[cnt] - k - 1);
            int rs = min(r, R[i]);
            if (rs >= ls)
                ans += rs - ls + 1;
        }
    }
    else {
        for (int i = cnt; i <= r; i++) {
            int ls = max(l, L[i]);
            int rs = min(i - (a[cnt] - k) + 1, cnt);
            if (rs >= ls) ans += rs - ls + 1;
        }
    }
    return ans + dfs(l, cnt - 1) + dfs(cnt + 1, r);
}
int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        scanf("%d%d", &n, &k);
        for (int i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        Log[0] = -1;
        for (int i = 1; i <= n; i++) Log[i] = Log[i >> 1] + 1;
        for (int i = 1; i <= n; i++)
            dp[i][0] = i;
        for (int j = 1; (1 << j) <= n; j++) {
            for (int i = 1; i + (1 << j) - 1 <= n; i++) {
                if (a[dp[i][j - 1]] > a[dp[i + (1 << (j - 1))][j - 1]])
                    dp[i][j] = dp[i][j - 1];
                else
                    dp[i][j] = dp[i + (1 << (j - 1))][j - 1];
            }
        }
        for (int i = 0; i <= n; i++)pos[i] = 0;
        R[n + 1] = n;
        for (int i = n; i >= 1; i--) {
            if (pos[a[i]]) {
                R[i] = pos[a[i]] - 1;
                pos[a[i]] = i;
            }
            else {
                pos[a[i]] = i;
                R[i] = n;
            }
            R[i] = min(R[i], R[i + 1]);
        }
        for (int i = 0; i <= n; i++)pos[i] = 0;
        for (int i = 1; i <= n; i++) {
            if (pos[a[i]]) {
                L[i] = pos[a[i]] + 1;
                pos[a[i]] = i;
            }
            else {
                pos[a[i]] = i;
                L[i] = 1;
            }
            L[i] = max(L[i], L[i - 1]);
        }
        printf("%lld
", dfs(1, n));
    }
}