[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 + ;
 int a[maxn], pos[maxn], L[maxn], R[maxn], Log[maxn];
 int n, k;
 int dp[maxn][];
 int query(int l, int r) {
     int k = Log[r - l + ];
     if (a[dp[l][k]] > a[dp[r - ( << k) + ][k]])
         return dp[l][k];
     else
         return dp[r - ( << k) + ][k];
 }
 ll dfs(int l, int r) {
     if (l == r)return a[l] -  <= k;
     if (l > r)return ;
     int cnt = query(l, r);
     ll ans = ;
     if (cnt - l < r - cnt) {
         for (int i = l; i <= cnt; i++) {
             int ls = max(cnt, i + a[cnt] - k - );
             int rs = min(r, R[i]);
             if (rs >= ls)
                 ans += rs - ls + ;
         }
     }
     else {
         for (int i = cnt; i <= r; i++) {
             int ls = max(l, L[i]);
             int rs = min(i - (a[cnt] - k) + , cnt);
             if (rs >= ls) ans += rs - ls + ;
         }
     }
     return ans + dfs(l, cnt - ) + dfs(cnt + , r);
 }
 int main() {
     int t;
     scanf("%d", &t);
     while (t--) {
         scanf("%d%d", &n, &k);
         for (int i = ; i <= n; i++)
             scanf("%d", &a[i]);
         Log[] = -;
         for (int i = ; i <= n; i++) Log[i] = Log[i >> ] + ;
         for (int i = ; i <= n; i++)
             dp[i][] = i;
         for (int j = ; ( << j) <= n; j++) {
             for (int i = ; i + ( << j) -  <= n; i++) {
                 if (a[dp[i][j - ]] > a[dp[i + ( << (j - ))][j - ]])
                     dp[i][j] = dp[i][j - ];
                 else
                     dp[i][j] = dp[i + ( << (j - ))][j - ];
             }
         }
         for (int i = ; i <= n; i++)pos[i] = ;
         R[n + ] = n;
         for (int i = n; i >= ; i--) {
             if (pos[a[i]]) {
                 R[i] = pos[a[i]] - ;
                 pos[a[i]] = i;
             }
             else {
                 pos[a[i]] = i;
                 R[i] = n;
             }
             R[i] = min(R[i], R[i + ]);
         }
         for (int i = ; i <= n; i++)pos[i] = ;
         for (int i = ; i <= n; i++) {
             if (pos[a[i]]) {
                 L[i] = pos[a[i]] + ;
                 pos[a[i]] = i;
             }
             else {
                 pos[a[i]] = i;
                 L[i] = ;
             }
             L[i] = max(L[i], L[i - ]);
         }
         printf("%lld\n", dfs(, n));
     }
 }