https://lydsy.com/JudgeOnline/problem.php?id=4826
分析：
莫队+单调栈+st表。
考虑如何O(1)加入一个点，删除一个点，类似bzoj4540。然后就可以莫队了。复杂度$O(n\sqrt n)$
代码：
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #include<iostream> 5 #include<cmath> 6 #include<cctype> 7 #include<set> 8 #include<queue> 9 #include<vector> 10 #include<map> 11 using namespace std; 12 typedef long long LL; 13 14 char buf[100000], *p1 = buf, *p2 = buf; 15 #define nc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++) 16 inline int read() { 17 int x=0,f=1;char ch=nc();for(;!isdigit(ch);ch=nc())if(ch=='-')f=-1; 18 for(;isdigit(ch);ch=nc())x=x*10+ch-'0';return x*f; 19 } 20 21 const int N = 200005; 22 struct Edge{ 23 int l, r, bel, id; 24 bool operator < (const Edge &A) const { 25 return bel == A.bel ? r < A.r : bel < A.bel; 26 } 27 }Q[N]; 28 int cnt[N], a[N], f[N][22], Log[N]; 29 int sum1R[N], sum1L[N], sum2R[N], sum2L[N], q[N], mxL[N], mxR[N]; 30 int P1, P2, L, R, n, m; 31 LL ans[N], Ans = 0; 32 33 int getmax(int l,int r) { 34 int k = Log[r - l + 1]; 35 int p1 = f[l][k], p2 = f[r - (1 << k) + 1][k]; 36 return a[p1] > a[p2] ? p1 : p2; 37 } 38 39 void updR(int x,int f) { 40 if (L >= R) { Ans = 0; return ; } 41 int p = getmax(L, R - 1), p2 = max(mxL[x], p); 42 Ans += (sum1L[R - 1] - sum1L[p2] + 1) * P1 * f; 43 Ans += (sum1L[p2] - sum1L[p]) * P2 * f; 44 Ans += (sum2L[R - 1] - sum2L[p2]) * P2 * f; 45 if (L > mxL[x]) Ans += (p2 - L) * P2 * f; 46 } 47 48 void updL(int x,int f) { 49 if (L >= R) { Ans = 0; return ; } 50 int p = getmax(L + 1, R), p2 = min(mxR[x], p); 51 Ans += (sum1R[L + 1] - sum1R[p2] + 1) * P1 * f; 52 Ans += (sum1R[p2] - sum1R[p]) * P2 * f; 53 Ans += (sum2R[L + 1] - sum2R[p2]) * P2 * f; 54 if (R < mxR[x]) Ans += (R - p2) * P2 * f; 55 } 56 57 int main() { 58 n = read(), m = read();P1 = read(), P2 = read(); 59 int B = sqrt(n); 60 for (int i = 1; i <= n; ++i) a[i] = read(), f[i][0] = i; 61 for (int j = 1; (1 << j) <= n; ++j) { 62 for (int i = 1; i + (1 << j) - 1 <= n; ++i) { 63 int p1 = f[i][j - 1], p2 = f[i + (1 << (j - 1))][j - 1]; 64 f[i][j] = a[p1] > a[p2] ? p1 : p2; 65 } 66 } 67 Log[0] = -1; 68 for (int i = 1; i <= n; ++i) Log[i] = Log[i >> 1] + 1; 69 a[0] = a[n + 1] = 1e9; 70 int l = 1, r = 1; q[1] = 0; 71 for (int i = 1; i <= n; ++i) { 72 while (l <= r && a[i] > a[q[r]]) r --; 73 q[++r] = i; 74 int j = q[r - 1]; mxL[i] = j; 75 sum2L[i] = sum2L[j] + (i - j - 1); sum1L[i] = sum1L[j] + 1; 76 } 77 l = 1, r = 1; q[1] = n + 1; 78 for (int i = n; i >= 1; --i) { 79 while (l <= r && a[i] > a[q[r]]) r --; 80 q[++r] = i; 81 int j = q[r - 1]; mxR[i] = j; 82 sum2R[i] = sum2R[j] + (j - i - 1); sum1R[i] = sum1R[j] + 1; 83 } 84 for (int i = 1; i <= m; ++i) { 85 Q[i].l = read(), Q[i].r = read(), Q[i].bel = (Q[i].l - 1) / B + 1, Q[i].id = i; 86 } 87 sort(Q + 1, Q + m + 1); 88 L = 1, R = 0; 89 for (int i = 1; i <= m; ++i) { 90 while (L > Q[i].l) L --, updL(L, 1); 91 while (R < Q[i].r) R ++, updR(R, 1); 92 while (L < Q[i].l) updL(L, -1), L ++; 93 while (R > Q[i].r) updR(R, -1), R --; 94 ans[Q[i].id] = Ans; 95 } 96 for (int i = 1; i <= m; ++i) printf("%lld\n", ans[i]); 97 return 0; 98 }