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 MBSubmit: 284 Solved: 108
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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。
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
