BZOJ 3277: 串/ BZOJ 3473: 字符串 ( 后缀数组 + RMQ + 二分 )

CF原题(http://codeforces.com/blog/entry/4849, 204E), CF的解法是O(Nlog^2N)的..记某个字符串以第i位开头的字符串对答案的贡献f(i), 那么f(i-1)>=f(i)-1, 然后就是O(NlogN)了, 但是速度垫底了....被后缀自动机完爆了...

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#include<cmath>

#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

typedef long long ll;

#define b(x) (1 << (x))

const int maxn = 200009;

int Sa[maxn], Rank[maxn], Height[maxn], S[maxn], cnt[maxn];

int L[maxn], R[maxn], Id[maxn]; // [L, R)

int buf[20], Pos[maxn], mn[20][maxn], K, N, n;

char str[maxn];

void Init() {

n = 0;

scanf("%d%d", &N, &K);

for(int i = 0; i < N; i++) {

scanf("%s", str);

L[i] = n;

for(int j = 0, g = strlen(str); j < g; j++) {

Id[n] = i;

S[n++] = str[j] - 'a' + 1;

}

R[i] = n;

Id[n] = N;

S[n++] = 0;

}

void Build() {

int m = 30, *x = Height, *y = Rank;

for(int i = 0; i < m; i++) cnt[i] = 0;

for(int i = 0; i < n; i++) cnt[x[i] = S[i]]++;

for(int i = 1; i < m; i++) cnt[i] += cnt[i - 1];

for(int i = 0; i < n; i++) Sa[--cnt[x[i]]] = i;

for(int k = 1, p = 0; k <= n; k <<= 1, p = 0) {

for(int i = n - k; i < n; i++) y[p++] = i;

for(int i = 0; i < n; i++)

if(Sa[i] >= k) y[p++] = Sa[i] - k;

for(int i = 0; i < m; i++) cnt[i] = 0;

for(int i = 0; i < n; i++) cnt[x[y[i]]]++;

for(int i = 1; i < m; i++) cnt[i] += cnt[i - 1];

for(int i = n; i--; ) Sa[--cnt[x[y[i]]]] = y[i];

swap(x, y);

p = 1;

x[Sa[0]] = 0;

for(int i = 1; i < n; i++) {

if(y[Sa[i]] != y[Sa[i - 1]] || y[Sa[i] + k] != y[Sa[i - 1] + k]) p++;

x[Sa[i]] = p - 1;

}

if(p >= n) break;

m = p;

}

for(int i = 0; i < n; i++) Rank[Sa[i]] = i;

Height[0] = 0;

for(int i = 0, h = 0; i < n; i++) if(Rank[i]) {

if(h) h--;

while(S[i + h] == S[Sa[Rank[i] - 1] + h]) h++;

Height[Rank[i]] = h;

}

void Init_RMQ() {

for(int i = 0; i < n; i++) mn[0][i] = Height[i];

for(int i = 1; b(i) <= n; i++)

for(int j = 0; j + b(i) <= n; j++)

mn[i][j] = min(mn[i - 1][j], mn[i - 1][j + b(i - 1)]);

}

inline int RMQ(int l, int r) {

if(r < l) return maxn;

int t = log2(r - l + 1);

return min(mn[t][l], mn[t][r - b(t) + 1]);

}

bool chk(int x, int len) {

int _L, _R;

if(Height[x] >= len) {

int l = 1, r = x;

while(l <= r) {

int m = (l + r) >> 1;

if(RMQ(m, x) >= len)

_L = m - 1, r = m - 1;

else

l = m + 1;

}

} else

_L = x;

if(x + 1 < n && Height[x + 1] >= len) {

int l = x, r = n - 1;

while(l <= r) {

int m = (l + r) >> 1;

if(RMQ(x + 1, m) >= len)

_R = m, l = m + 1;

else

r = m - 1;

}

} else

_R = x;

return ~Pos[_R] && Pos[_R] >= _L;

}

void Work() {

Build();

for(int i = 0; i < N; i++) cnt[i] = 0;

cnt[N] = N;

for(int i = 0, p = 0, tot = 0; i < n; i++) {

if(!cnt[Id[Sa[i]]]) tot++;

cnt[Id[Sa[i]]]++;

while(tot > K || (tot == K && cnt[Id[Sa[p]]] > 1))

if(!--cnt[Id[Sa[p++]]]) tot--;

if(tot >= K) {

Pos[i] = p;

} else

Pos[i] = -1;

}

Init_RMQ();

for(int i = 0; i < N; i++) {

ll tot = 0;

for(int j = L[i], Last = 0; j < R[i]; j++) {

int ans = max(Last - 1, 0);

while(ans < R[i] - j && chk(Rank[j], ans + 1)) ans++;

Last = ans;

tot += ans;

}

printf("%I64d ", tot);

}

puts("");

}

int main() {

Init();

Work();

return 0;

}

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3277: 串

Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 284 Solved: 108
[Submit][Status][Discuss]

Description

字符串是oi界常考的问题。现在给定你n个字符串，询问每个字符串有多少子串（不包括空串）是所有n个字符串中至少k个字符串的子串（注意包括本身）。

Input

第一行两个整数n，k。
　　接下来n行每行一个字符串。

Output

输出一行n个整数，第i个整数表示第i个字符串的答案。

Sample Input

3 1
abc
a
ab

Sample Output

6 1 3

HINT

对于100%的数据，n,k,l<=100000

Source

后缀数组

巴特西

BZOJ 3277: 串/ BZOJ 3473: 字符串 ( 后缀数组 + RMQ + 二分 )

3277: 串

Description

Input

Output

Sample Input

Sample Output

HINT

Source

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