Description
XX在进行字符串研究的时候,遇到了一个十分棘手的问题。
在这个问题中,给定一个字符串S,与一个整数K,定义S的子串T=S(i, j)是关于第K位的识别子串,满足以下两个条件:
1、i≤K≤j。
2、子串T只在S中出现过一次。
例如,S="banana",K=5,则关于第K位的识别子串有"nana","anan","anana","nan","banan"和"banana"。
现在,给定S,XX希望知道对于S的每一位,最短的识别子串长度是多少,请你来帮助他。
Input
仅一行,输入长度为N的字符串S。
Output
输出N行,每行一个整数,第i行的整数表示对于第i位的最短识别子串长度。
Sample Input
agoodcookcooksgoodfood
Sample Output
1
2
3
3
2
2
3
3
2
2
3
3
2
1
2
3
3
2
1
2
3
4
HINT
N<=5*10^5
首先发现可以按照每个后缀统计贡献
然后直接把每个后缀和前后排名的串lcp的max:len求出来,这些值是不能更新的
然后对于$[i,i+len] (用len+1更新,对于)[i+len,n]$用一个等差数列更新min
等差数列维护直接标记永久化就可以了
然后注意特判是不是没有合法的解
#include<bits/stdc++.h>
using namespace std;
typedef pair<int, int> pi;
const int N = 5e5 + 10;
const int INF_of_int = 1e9;
int typ1[N << 2], typ2[N << 2];
#define LD (t << 1)
#define RD (t << 1 | 1)
void build(int t, int l, int r) {
typ1[t] = INF_of_int;
typ2[t] = -INF_of_int;
if (l == r) return;
int mid = (l + r) >> 1;
build(LD, l, mid);
build(RD, mid + 1, r);
}
void modify(int t, int l, int r, int ql, int qr, int typ, int val) {
if (ql > qr) return;
if (ql <= l && r <= qr) {
if (typ == 1) {
typ1[t] = min(typ1[t], val);
} else {
typ2[t] = max(typ2[t], val);
}
return;
}
int mid = (l + r) >> 1;
if (qr <= mid) modify(LD, l, mid, ql, qr, typ, val);
else if (ql > mid) modify(RD, mid + 1, r, ql, qr, typ, val);
else {
modify(LD, l, mid, ql, mid, typ, val);
modify(RD, mid + 1, r, mid + 1, qr, typ, val);
}
}
void output(int t, int l, int r, int val, int pos) {
if (typ1[t]) val = min(val, typ1[t]);
if (typ2[t]) val = min(val, pos - typ2[t] + 1);
if (l == r) {
printf("%d
", val);
return;
}
int mid = (l + r) >> 1;
if (pos <= mid) output(LD, l, mid, val, pos);
else output(RD, mid + 1, r, val, pos);
}
struct Suffix_Array {
int s[N], n, m;
int c[N], x[N], y[N];
int sa[N], rank[N], height[N];
void init(int len, char *c) {
n = len, m = 0;
for (int i = 1; i <= n; i++) {
s[i] = c[i];
m = max(m, s[i]);
}
}
void radix_sort() {
for (int i = 1; i <= m; i++) c[i] = 0;
for (int i = 1; i <= n; i++) c[x[y[i]]]++;
for (int i = 1; i <= m; i++) c[i] += c[i - 1];
for (int i = n; i >= 1; i--) sa[c[x[y[i]]]--] = y[i];
}
void buildsa() {
for (int i = 1; i <= n; i++) x[i] = s[i], y[i] = i;
radix_sort();
int now;
for (int k = 1; k <= n; k <<= 1) {
now = 0;
for (int i = n - k + 1; i <= n; i++) y[++now] = i;
for (int i = 1; i <= n; i++) if (sa[i] > k) y[++now] = sa[i] - k;
radix_sort();
y[sa[1]] = now = 1;
for (int i = 2; i <= n; i++) y[sa[i]] = (x[sa[i]] == x[sa[i - 1]] && x[sa[i] + k] == x[sa[i - 1] + k]) ? now : ++now;
swap(x, y);
if (now == n) break;
m = now;
}
}
void buildrank() {
for (int i = 1; i <= n; i++) rank[sa[i]] = i;
}
void buildheight() {
for (int i = 1; i <= n; i++) {
int k = max(height[rank[i - 1]] - 1, 0);
for (; s[i + k] == s[sa[rank[i] - 1] + k]; k++);
height[rank[i]] = k;
}
}
void build(int len, char *c) {
init(len, c);
buildsa();
buildrank();
buildheight();
}
void solve() {
for (int i = 1; i <= n; i++) {
int len = max(height[rank[i]], height[rank[i] + 1]);
if (i + len > n) continue;
modify(1, 1, n, i, i + len, 1, len + 1);
modify(1, 1, n, i + len, n, 2, i);
}
}
} Sa;
int len;
char s[N];
int main() {
#ifdef dream_maker
freopen("input.txt", "r", stdin);
#endif
scanf("%s", s + 1);
len = strlen(s + 1);
Sa.build(len, s);
build(1, 1, len);
Sa.solve();
for (int i = 1; i <= len; i++) output(1, 1, len, len, i);
return 0;
}