BZOJ 3230: 相似子串( RMQ + 后缀数组 + 二分 )

二分查找求出k大串, 然后正反做后缀数组, RMQ求LCP, 时间复杂度O(NlogN+logN)

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#include<cstdio>

#include<algorithm>

#include<cstring>

#include<cctype>

 

using namespace std;

 

typedef long long ll;

 

const int maxlog = 20;

const int maxn = 100009;

 

int N, Q;

 

inline int readint() {

char c = getchar();

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

int ret = 0;

for(; isdigit(c); c = getchar())

ret = ret * 10 + c - '0';

return ret;

}

 

inline ll readll() {

char c = getchar();

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

ll ret = 0;

for(; isdigit(c); c = getchar())

ret = ret * 10 + c - '0';

return ret;

}

 

struct SA {

int Sa[maxn], Rank[maxn], Height[maxn], cnt[maxn], RMQ[maxlog][maxn];

ll Sm[maxn];

char S[maxn];

#define Cmp(a, b) ((y[a] == y[b]) && (y[a + k] == y[b + k]))

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

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);

p = 1;

x[Sa[0]] = 0;

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

x[Sa[i]] = Cmp(Sa[i], Sa[i - 1]) ? p - 1 : p++;

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 Query_Init() {

Sm[0] = N - Sa[0] - 1;

for(int i = 1; i < N; i++) Sm[i] = Sm[i - 1] + N - Sa[i] - Height[i] - 1;

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)]);

}

int LCP(int x, int y) {

x = Rank[x], y = Rank[y];

if(x == y) return N;

if(x > y) swap(x, y);

x++;

int Log = 0;

while(b(Log) <= y - x + 1) Log++;

Log--;

return min(RMQ[Log][x], RMQ[Log][y - b(Log) + 1]);

}

pair<int, int> Get(ll v) {

int p = lower_bound(Sm, Sm + N, v) - Sm;

if(p >= N) return make_pair(-1, -1);

if(p) v -= Sm[p - 1];

return make_pair(Sa[p], Sa[p] + v + Height[p] - 1);

}

} A, B;

 

void Init() {

N = readint(), Q = readint();

char c = getchar();

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

A.S[0] = B.S[N - 1] = c;

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

A.S[i] = B.S[N - i - 1] = getchar();

A.S[N] = B.S[N] = '$';

N++;

}

 

#define L(x) x.first

#define R(x) x.second

 

void Work() {

A.Build(), A.Query_Init();

B.Build(), B.Query_Init();

ll l, r;

while(Q--) {

l = readll(), r = readll();

pair<int, int> L = A.Get(l), R = A.Get(r);

if(L(L) == -1 || L(R) == -1) {

puts("-1");

} else {

int mn = min(R(L) - L(L) + 1, R(R) - L(R) + 1);

int a = min(A.LCP(L(L), L(R)), mn);

int b = min(B.LCP(N - R(L) - 2, N - R(R) - 2), mn);

printf("%lld\n", ll(a) * a + ll(b) * b);

}

}

}

 

int main() {

Init();

Work();

return 0;

}

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3230: 相似子串

Time Limit: 20 Sec  Memory Limit: 128 MB
Submit: 1186  Solved: 282
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Description

Input

输入第1行，包含3个整数N，Q。Q代表询问组数。
第2行是字符串S。
接下来Q行，每行两个整数i和j。（1≤i≤j）。

Output

输出共Q行，每行一个数表示每组询问的答案。如果不存在第i个子串或第j个子串，则输出-1。

Sample Input

5 3
ababa
3 5
5 9
8 10

Sample Output

18
16
-1

HINT

样例解释

第1组询问：两个子串是“aba”,“ababa”。f = 32 + 32 = 18。

第2组询问：两个子串是“ababa”,“baba”。f = 02 + 42 = 16。

第3组询问：不存在第10个子串。输出-1。

数据范围

N≤100000，Q≤100000，字符串只由小写字母'a'~'z'组成

Source

后缀数组+二分+RMQ