BZOJ 2780: [Spoj]8093 Sevenk Love Oimaster( 后缀数组 + 二分 + RMQ + 树状数组 )

全部串起来做SA, 在按字典序排序的后缀中, 包含每个询问串必定是1段连续的区间, 对每个询问串s二分+RMQ求出包含s的区间. 然后就是求区间的不同的数的个数(经典问题), sort queries + BIT 就行了.时间复杂度O(N log N). 速度垫底了QAQ 你们都会SAM。。。。

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#include<cmath>

#include<cstdio>

#include<cctype>

#include<cstring>

#include<algorithm>

using namespace std;

#define b(i) (1 << (i))

const int maxL = 540009;

const int maxQ = 60009;

char S[maxL], str[maxL];

int N, n, q, Id[maxL], qL[maxQ], qR[maxQ], L[maxQ], R[maxQ];

int Rank[maxL], Height[maxL], Sa[maxL], cnt[maxL];

int RMQ[20][maxL], r[maxQ], ans[maxQ];

void Build() {

int m = ‘z‘ + 1, *x = Rank, *y = Height;

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 = 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);

x[Sa[0]] = 0;

p = 1;

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((m = p) >= N) break;

}

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

Height[0] = Height[N] = 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++)

RMQ[0][i] = Height[i];

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

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

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

}

inline int LCP(int l, int r) {

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

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

}

void calc(int &L, int &R, int p, int len) {

int _l, _r;

p = Rank[p];

if(Height[p] >= len) {

_l = 0, _r = p - 1;

while(_l <= _r) {

int m = (_l + _r) >> 1;

if(LCP(m + 1, p) >= len)

L = m, _r = m - 1;

else

_l = m + 1;

}

} else

L = p;

if(Height[p + 1] >= len) {

_l = p + 1, _r = N - 1;

while(_l <= _r) {

int m = (_l + _r) >> 1;

if(LCP(p + 1, m) >= len)

R = m, _l = m + 1;

else

_r = m - 1;

}

} else

R = p;

}

struct Link {

int p;

Link* n;

} pool[maxL], *pt = pool, *H[maxL];

inline void AddL(int v, int p) {

pt->p = p, pt->n = H[v], H[v] = pt++;

}

int B[maxL];

inline void Modify(int p, int v) {

if(!p) return;

for(; p <= N; p += p & -p) B[p] += v;

}

inline int Sum(int p) {

int ret = 0;

for(; p; p -= p & -p) ret += B[p];

return ret;

}

inline bool Cmp(const int &l, const int &r) {

return qL[l] < qL[r];

}

void Work() {

Build();

Init_RMQ();

memset(B, 0, sizeof B);

for(int i = N; i--; )

if(Id[Sa[i]] >= 0) AddL(Id[Sa[i]], i);

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

Modify(H[i]->p + 1, 1);

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

calc(qL[r[i] = i], qR[i], L[i], R[i] - L[i]);

sort(r, r + q, Cmp);

int c = 0;

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

while(qL[r[c]] == i) {

ans[r[c]] = Sum(qR[r[c]] + 1) - Sum(qL[r[c]]);

if(++c >= q) break;

}

if(c >= q) break;

Modify(i + 1, -1);

if(H[Id[Sa[i]]]) {

H[Id[Sa[i]]] = H[Id[Sa[i]]]->n;

if(H[Id[Sa[i]]])

Modify(H[Id[Sa[i]]]->p + 1, 1);

}

}

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

printf("%d\n", ans[i]);

}

inline int getstr() {

char c = getchar();

for(; !islower(c); c = getchar());

int len = 0;

for(; islower(c); c = getchar())

str[len++] = c;

return len;

}

void Init() {

scanf("%d%d", &n, &q);

N = 0;

int len;

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

len = getstr();

for(int j = 0; j < len; j++) {

Id[N] = i;

S[N++] = str[j];

}

Id[N] = -1;

S[N++] = ‘$‘;

}

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

len = getstr();

L[i] = N;

for(int j = 0; j < len; j++) {

Id[N] = -1;

S[N++] = str[j];

}

R[i] = N;

Id[N] = -1;

S[N++] = ‘$‘;

}

S[N - 1] = 0;

}

int main() {

Init();

Work();

return 0;

}

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2780: [Spoj]8093 Sevenk Love Oimaster

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 581  Solved: 188
[Submit][Status][Discuss]

Description

Oimaster and sevenk love each other.

Input

Output

Sample Input

3 3
abcabcabc
aaa
aafe
abc
a
ca

Sample Output

1
3
1

HINT

Source