3238: [Ahoi2013]差异

Time Limit: 20 Sec Memory Limit: 512 MB
Submit: 2106 Solved: 953
[Submit][Status][Discuss]

Description

Input

一行，一个字符串S

Output

一行，一个整数，表示所求值

Sample Input

cacao

Sample Output

HINT

2<=N<=500000,S由小写英文字母组成

Source

[Submit][Status][Discuss]

构造一个后缀数组，两个后缀的lcp就是在这个数组走过去时取min，单调队列瞎搞即可 写这个真的想吐。。。

#include<iostream>
#include<cstdio>
#include<queue>
#include<vector>
#include<bitset>
#include<algorithm>
#include<cstring>
#include<map>
#include<stack>
#include<set>
#include<cmath>
#include<ext/pb_ds/priority_queue.hpp>
using namespace std;const int maxn = 5E5 + 50;
typedef long long LL;int n,m,co[maxn],t[maxn],t2[maxn],height[maxn],l[maxn],r[maxn],q[maxn],rank[maxn],sa[maxn];
char ch[maxn];
LL ans = 0;void Getsa()
{int *x = t,*y = t2; m = 26;for (int i = 1; i <= n; i++) ++co[x[i] = ch[i]];for (int i = 2; i <= m; i++) co[i] += co[i-1]; for (int i = n; i; i--) sa[co[x[i]]--] = i; for (int k = 1; k < n; k <<= 1) {int p = 0;for (int i = n; i > n - k; i--) y[++p] = i;for (int i = 1; i <= n; i++) if (sa[i] - k > 0) y[++p] = sa[i] - k;for (int i = 1; i <= m; i++) co[i] = 0;for (int i = 1; i <= n; i++) ++co[x[y[i]]];for (int i = 2; i <= m; i++) co[i] += co[i-1];for (int i = n; i; i--) sa[co[x[y[i]]]--] = y[i];swap(x,y); x[sa[1]] = (p = 1);for (int i = 2; i <= n; i++)x[sa[i]] = y[sa[i]] == y[sa[i-1]] && y[sa[i]+k] == y[sa[i-1]+k]?p:++p;if (p == n) break;m = p; }
}void Rank_and_Height()
{for (int i = 1; i <= n; i++) rank[sa[i]] = i;int k = 0;for (int i = 1; i <= n; i++) {if (k) --k;int j = sa[rank[i]-1];while (ch[i + k] == ch[j + k]) ++k;height[rank[i]] = k;}
}int main()
{#ifdef DMCfreopen("DMC.txt","r",stdin);#endifscanf("%s",ch + 1); n = strlen(ch + 1);for (int i = 1; i <= n; i++) ch[i] = ch[i] - 'a' + 1;Getsa(); Rank_and_Height();int tail = 0; q[tail = 1] = 0;for (int i = 1; i <= n + 1; i++) {while (tail && height[q[tail]] > height[i]) r[q[tail--]] = i-1;q[++tail] = i;}while (tail) r[q[tail--]] = n;q[tail = 1] = n + 1;for (int i = n; i >= 0; i--) {while (tail && height[q[tail]] >= height[i]) l[q[tail--]] = i+1;q[++tail] = i;}while (tail) l[q[tail--]] = 1;for (int i = 2; i <= n; i++) ans -= 2LL*height[i]*(i-l[i]+1LL)*(r[i]-i+1LL);LL K = 1LL*(1+n)*n/2LL;cout << K*1LL*(n-1LL) + ans;return 0;
}

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3238: [Ahoi2013]差异

3238: [Ahoi2013]差异

Description

Input

Output

Sample Input

Sample Output

HINT

Source

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