很久没写过这东西了，复习一波。
3238: [Ahoi2013]差异
单调栈维护height数组，由于height是递增的，所以维护单调栈中维护每个height出现的次数。（还可以两遍单调栈求一个点是最小值的区间）
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 inline int read() { 15 int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; 16 for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f; 17 } 18 19 const int N = 500005; 20 21 char s[N]; 22 int t1[N], t2[N], c[N], sa[N], rnk[N], height[N], m = 130, n; 23 LL sk[N], cnt[N]; 24 25 void getsa() { 26 int *x = t1, *y = t2, i, p; 27 for (i = 1; i <= m; ++i) c[i] = 0; 28 for (i = 1; i <= n; ++i) x[i] = s[i], c[x[i]] ++; 29 for (i = 1; i <= m; ++i) c[i] += c[i - 1]; 30 for (i = 1; i <= n; ++i) sa[c[x[i]]--] = i; 31 for (int k = 1; k <= n; k <<= 1) { 32 p = 0; 33 for (i = n - k + 1; i <= n; ++i) y[++p] = i; 34 for (i = 1; i <= n; ++i) if (sa[i] > k) y[++p] = sa[i] - k; 35 for (i = 1; i <= m; ++i) c[i] = 0; 36 for (i = 1; i <= n; ++i) c[x[y[i]]] ++; 37 for (i = 1; i <= m; ++i) c[i] += c[i - 1]; 38 for (i = n; i >= 1; --i) sa[c[x[y[i]]]--] = y[i]; 39 swap(x, y); 40 p = 2; 41 x[sa[1]] = 1; 42 for (i = 2; i <= n; ++i) 43 x[sa[i]] = (y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k]) ? p - 1 : p ++; 44 if (p > n) break; 45 m = p; 46 } 47 } 48 void getheight() { 49 for (int i = 1; i <= n; ++i) rnk[sa[i]] = i; 50 int k = 0; 51 height[1] = 0; 52 for (int i = 1; i <= n; ++i) { 53 if (rnk[i] == 1) continue; 54 if (k) k --; 55 int j = sa[rnk[i] - 1]; 56 while (i + k <= n && j + k <= n && s[i + k] == s[j + k]) k ++; 57 height[rnk[i]] = k; 58 } 59 } 60 int main() { 61 scanf("%s", s + 1); 62 n = strlen(s + 1); 63 getsa(); 64 getheight(); 65 LL ans = 0; 66 for (int i = 1; i <= n; ++i) ans += 1ll * (n - 1) * i; 67 int top = 0;LL now = 0; 68 for (int i = 2; i <= n; ++i) { 69 LL tmp = 1; 70 while (top && sk[top] >= height[i]) { 71 now -= 1ll * cnt[top] * sk[top]; tmp += cnt[top]; top --; 72 } 73 sk[++top] = height[i]; cnt[top] = tmp; now += 1ll * height[i] * tmp; 74 ans -= now * 2; 75 } 76 cout << ans; 77 return 0; 78 }