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[AHOI2013] 差异 题解

[AHOI2013] 差异 题解

题面

题目链接

解法

使用后缀数组求出$height$数组(简写为$ht$),答案变成了

化简,得

求和符号里的用单调栈维护。

时间复杂度$O(n \log n)$。

AC代码

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/**
* @file: P4248.cpp
* @author: yaoxi-std
* @url: https://www.luogu.com.cn/problem/P4248
*/
// #pragma GCC optimize ("O2")
// #pragma GCC optimize ("Ofast", "inline", "-ffast-math")
// #pragma GCC target ("avx,sse2,sse3,sse4,mmx")
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define resetIO(x) \
freopen(#x ".in", "r", stdin), freopen(#x ".out", "w", stdout)
#define debug(fmt, ...) \
fprintf(stderr, "[%s:%d] " fmt "\n", __FILE__, __LINE__, ##__VA_ARGS__)
template <class _Tp>
inline _Tp& read(_Tp &x) {
bool sign = false;
char ch = getchar();
long double tmp = 1;
for (; !isdigit(ch); ch = getchar())
sign |= (ch == '-');
for (x = 0; isdigit(ch); ch = getchar())
x = x * 10 + (ch ^ 48);
if (ch == '.')
for (ch = getchar(); isdigit(ch); ch = getchar())
tmp /= 10.0, x += tmp * (ch ^ 48);
return sign ? (x = -x) : x;
}
template <class _Tp>
inline void write(_Tp x) {
if (x < 0)
putchar('-'), x = -x;
if (x > 9)
write(x / 10);
putchar((x % 10) ^ 48);
}
#define fi first
#define se second
using pii = pair<int, int>;
const int MAXN = 5e5 + 10;
const int INF = 0x3f3f3f3f3f3f3f3f;
struct suffix_array {
int n, sa[MAXN], rk[MAXN], tp[MAXN], ht[MAXN];
void radix_sort(int m) {
static int buk[MAXN];
for (int i = 0; i <= m; ++i)
buk[i] = 0;
for (int i = 1; i <= n; ++i)
buk[rk[i]]++;
for (int i = 1; i <= m; ++i)
buk[i] += buk[i - 1];
for (int i = n; i >= 1; --i)
sa[buk[rk[tp[i]]]--] = tp[i];
}
void init(char *s, int n) {
this->n = n;
int m = 100;
for (int i = 1; i <= n; ++i)
rk[i] = s[i] - 'a' + 1, tp[i] = i;
radix_sort(m);
for (int p = 0, w = 1; p < n; m = p, w <<= 1) {
p = 0;
for (int i = 1; i <= w; ++i)
tp[++p] = n - w + i;
for (int i = 1; i <= n; ++i)
if (sa[i] > w)
tp[++p] = sa[i] - w;
radix_sort(m);
copy(rk + 1, rk + n + 1, tp + 1);
rk[sa[1]] = p = 1;
for (int i = 2; i <= n; ++i) {
if (tp[sa[i - 1]] == tp[sa[i]] && tp[sa[i - 1] + w] == tp[sa[i] + w])
rk[sa[i]] = p;
else
rk[sa[i]] = ++p;
}
}
for (int i = 1, k = 0; i <= n; ++i) {
if (k)
k--;
while (s[i + k] == s[sa[rk[i] - 1] + k])
k++;
ht[i] = k;
}
}
};
int n, top;
pii sta[MAXN];
char s[MAXN];
suffix_array sa;
signed main() {
scanf("%s", s + 1);
n = strlen(s + 1);
sa.init(s, n);
int ans = 0, sum = 0;
sta[top = 1] = {0, 1};
for (int i = 2; i <= n; ++i) {
int p = sa.sa[i];
while (top && sta[top].fi >= sa.ht[p])
sum -= sta[top].fi * (sta[top].se - sta[top - 1].se), --top;
sta[++top] = {sa.ht[p], i}, sum += sta[top].fi * (sta[top].se - sta[top - 1].se);
ans += sum;
}
ans = (n * (n + 1) * (n - 1)) / 2 - ans * 2;
write(ans), putchar('\n');
return 0;
}