BZOJ3230 相似子串 后缀数组

Posted 2020-11-05 Mychael

题目链接

BZOJ3230
权限题

题解

后缀数组基础题
询问第K大不同子串和正反lcp长度
如果您RE了，您就要知道询问的输入会爆LL

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 400005,maxm = 100005,INF = 1000000000;
inline LL read(){
    LL out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == ‘-‘) flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
int n,m,bac[maxn],t1[maxn],t2[maxn],bin[50],Log[maxn];
LL sum[maxn];
struct SA{
    char s[maxn];
    int sa[maxn],rank[maxn],height[maxn],mn[maxn][18];
    void getsa(){
        int *x = t1,*y = t2; m = 255;
        for (int i = 0; i <= m; i++) bac[i] = 0;
        for (int i = 1; i <= n; i++) bac[x[i] = s[i]]++;
        for (int i = 1; i <= m; i++) bac[i] += bac[i - 1];
        for (int i = n; i; i--) sa[bac[x[i]]--] = i;
        for (int k = 1; k <= n; k <<= 1){
            int p = 0;
            for (int i = n - k + 1; i <= n; i++) y[++p] = i;
            for (int i = 1; i <= n; i++) if (sa[i] - k > 0) y[++p] = sa[i] - k;
            for (int i = 0; i <= m; i++) bac[i] = 0;
            for (int i = 1; i <= n; i++) bac[x[y[i]]]++;
            for (int i = 1; i <= m; i++) bac[i] += bac[i - 1];
            for (int i = n; i; i--) sa[bac[x[y[i]]]--] = y[i];
            swap(x,y);
            x[sa[1]] = p = 1;
            for (int i = 2; i <= n; i++)
                x[sa[i]] = (y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k] ? p : ++p);
            if (p >= n) break;
            m = p;
        }
        for (int i = 1; i <= n; i++) rank[sa[i]] = i;
        for (int i = 1,k = 0; i <= n; i++){
            if (k) k--;
            int j = sa[rank[i] - 1];
            while (s[i + k] == s[j + k]) k++;
            height[rank[i]] = k;
        }
        for (int i = 1; i <= n; i++) mn[i][0] = height[i];
        for (int j = 1; j <= 17; j++)
            for (int i = 1; i <= n; i++){
                if (i + bin[j] - 1 > n) break;
                mn[i][j] = min(mn[i][j - 1],mn[i + bin[j - 1]][j - 1]);
            }
    }
    int lcp(int x,int y){
        if (x == y) return n - x + 1;
        int l = rank[x],r = rank[y];
        if (l > r) swap(l,r); l++;
        int t = Log[r - l + 1];
        return min(mn[l][t],mn[r - bin[t] + 1][t]);
    }
}L,R;
void init(){
    sum[1] = n - L.sa[1] + 1;
    for (int i = 2; i <= n; i++)
        sum[i] = sum[i - 1] + (n - L.sa[i] + 1) - L.height[i];
}
int q;
void solve(){
    LL f,a,b,x,y;
    int l1,r1,l2,r2;
    while (q--){
        x = read(); y = read();
        if (x > sum[n] || y > sum[n]) {puts("-1"); continue;}
        int p = lower_bound(sum + 1,sum + 1 + n,x) - sum;
        int q = lower_bound(sum + 1,sum + 1 + n,y) - sum;
        if (sum[p] == x) l1 = L.sa[p],r1 = n;
        else l1 = L.sa[p],r1 = n - (sum[p] - x);
        if (sum[q] == y) l2 = L.sa[q],r2 = n;
        else l2 = L.sa[q],r2 = n - (sum[q] - y);
        a = min(L.lcp(l1,l2),min(r1 - l1 + 1,r2 - l2 + 1));
        b = min(R.lcp(n - r1 + 1,n - r2 + 1),min(r1 - l1 + 1,r2 - l2 + 1));
        f = a * a + b * b;
        printf("%lld\n",f);
    }
}
int main(){
    bin[0] = 1; for (int i = 1; i <= 30; i++) bin[i] = bin[i - 1] << 1;
    Log[0] = -1; for (int i = 1; i <= 100000; i++) Log[i] = Log[i >> 1] + 1;
    n = read(); q = read();
    scanf("%s",L.s + 1);
    for (int i = 1; i <= n; i++) R.s[i] = L.s[n - i + 1];
    L.getsa(); R.getsa();
    init(); solve();
    return 0;
}