HDU 6194 string string string

source：2017 ACM/ICPC Asia Regional Shenyang Online

题意：给你n个字符串，问恰好出现k次的字符串有几种。

用后缀数组构造出height数组后，注意到对于一段连续的长度为k的区间，前缀长度的下界为max(h[L],h[R+1)，上界为min(h[L+1],h[L+2]...,h[R])，这里对上界的求法可以用RMQ维护，也可以用一个单调递增的双端队列维护。
注意k==1时特判。

代码：

#include<bits/stdc++.h>using namespace std;#define fi first#define se second#define pb push_back#define CLR(A, X) memset(A, X, sizeof(A))#define bitcount(X) __builtin_popcountll(X)typedef long long LL;typedef pair<int, int> PII;const double eps = 1e-10;const LL MOD = 1e9+7;const auto INF = 0x3f3f3f3f;int dcmp(double x) { if(fabs(x) < eps) return 0; return x<0?-1:1; }const int MAXN = 1e5+10;char str[MAXN];int s[MAXN], sa[MAXN], x[MAXN], y[MAXN], c[MAXN];int Rank[MAXN], h[MAXN];int q[MAXN];void get_sa(int m, int n) {    int i, k, p;    for(i = 0; i < m; i++) c[i] = 0;    for(i = 0; i < n; i++) c[x[i] = s[i]]++;    for(i = 1; i < m; i++) c[i] += c[i-1];    for(i = n-1; i >= 0; i--) sa[--c[x[i]]] = i;    for(k = 1, p = 1; k<=n && p<n; k <<= 1, m = p) {        for(i = n-k, p = 0; i < n; i++) y[p++] = i;        for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i]-k;        for(i = 0; i < m; i++) c[i] = 0;        for(i = 0; i < n; i++) c[x[y[i]]]++;        for(i = 1; i < m; i++) c[i] += c[i-1];        for(i = n-1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];        swap(x, y); p = 1; x[sa[0]] = 0;        for(i = 1; i < n; i++) {            int l = sa[i-1], r = sa[i];            if(y[l]==y[r] && y[l+k]==y[r+k]) x[sa[i]] = p-1;            else x[sa[i]] = p++;        }    }    for(i = 0; i < n; i++) Rank[sa[i]] = i;    for(i = 0, k = 0; i < n; i++) {        if(k) k--; if(!Rank[i]) continue;        int j = sa[Rank[i]-1];        while(s[j+k] == s[i+k]) k++;        h[Rank[i]] = k;    }}int main() {    int X, k;    scanf("%d", &X);    while(X--) {        scanf("%d%s", &k, str);        int n = strlen(str);        for(int i = 0; i < n; i++) s[i] = str[i]-'a'+1;        s[n] = h[0] = 0; h[++n] = 0;        get_sa(27, n);        int L = 0, R = -1; LL ans = 0;        if(k == 1) {            for(int i = 1; i < n; i++) ans += n-sa[i]-1-max(h[i], h[i+1]);            printf("%lld\n", ans);            continue;        }        for(int i = 2; i < n; i++) {            while(L<=R && h[q[R]]>h[i]) R--;            q[++R] = i;            while(i-q[L]+1 >= k) L++;            if(i >= k) {                int t1 = h[q[L]], t2 = max(h[i-k+1], h[i+1]);                ans += max(t1-t2, 0);            }        }        printf("%lld\n", ans);    }    return 0;}