• bzoj 3676 后缀自动机+马拉车+树上倍增


    思路:用马拉车把一个串中的回文串个数降到O(n)级别,然后每个串在后缀自动机上倍增找个数。

    #include<bits/stdc++.h>
    #define LL long long
    #define fi first
    #define se second
    #define mk make_pair
    #define PII pair<int, int>
    #define PLI pair<LL, int>
    #define ull unsigned long long
    using namespace std;
    
    const int N = 300000 + 7;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const int mod = 1e9 + 7;
    const double eps = 1e-8;
    const int base = 87;
    
    int n, m, p[N<<1];
    char s[N<<1];
    
    struct SuffixAutomaton {
        int last, cur, cnt, ch[N<<1][26], id[N<<1], fa[N<<1], dis[N<<1], sz[N<<1], c[N];
        int f[N<<1][20], pos[N<<1];
        SuffixAutomaton() {cur = cnt = 1;}
        void init() {
            for(int i = 1; i <= cnt; i++) {
                memset(ch[i], 0, sizeof(ch[i]));
                sz[i] = c[i] = dis[i] = fa[i] = 0;
            }
            cur = cnt = 1;
        }
        void extend(int c, int id) {
            last = cur; cur = ++cnt;
            int p = last; dis[cur] = id;
            for(; p && !ch[p][c]; p = fa[p]) ch[p][c] = cur;
            if(!p) fa[cur] = 1;
            else {
                int q = ch[p][c];
                if(dis[q] == dis[p]+1) fa[cur] = q;
                else {
                    int nt = ++cnt; dis[nt] = dis[p]+1;
                    memcpy(ch[nt], ch[q], sizeof(ch[q]));
                    fa[nt] = fa[q]; fa[q] = fa[cur] = nt;
                    for(; ch[p][c]==q; p=fa[p]) ch[p][c] = nt;
                }
            }
            sz[cur] = 1;
        }
        void getSize(int n) {
            for(int i = 1; i <= cnt; i++) c[dis[i]]++;
            for(int i = 1; i <= n; i++) c[i] += c[i-1];
            for(int i = cnt; i >= 1; i--) id[c[dis[i]]--] = i;
            for(int i = cnt; i >= 1; i--) {
                int p = id[i];
                sz[fa[p]] += sz[p];
            }
        }
        LL query(int p, int len) {
            for(int j = 19; j >= 0; j--) {
                if(f[p][j] && dis[f[p][j]] >= len) p = f[p][j];
            }
            return 1ll*len*sz[p];
        }
        void solve() {
            for(int i = 1, p = 1; i <= n; i++)
                p = ch[p][s[i]-'a'], pos[i] = p;
            for(int i = 1; i <= cnt; i++) f[i][0] = fa[i];
            for(int j = 1; j < 20; j++)
                for(int i = 1; i <= cnt; i++)
                    f[i][j] = f[f[i][j-1]][j-1];
    
            LL ans = 0;
            s[0] = '-', s[n+1] = '+';
            int mx = 0, id = 0;
            for(int i = 1; i <= n; i++) {
                if(mx > i) p[i] = min(mx-i, p[2*id-i]);
                else p[i]=1, ans = max(ans, query(pos[i], 1));
                while(s[i+p[i]]==s[i-p[i]]) p[i]++, ans = max(ans, query(pos[i+p[i]-1], 2*p[i]-1));
                if(i+p[i]>mx) mx = i+p[i], id = i;
            }
            mx = 0, id = 0;
            for(int i = 1; i <= n; i++) {
                if(mx > i) p[i] = min(mx-i, p[2*id-i]);
                else p[i] = 0;
                while(s[i+p[i]+1]==s[i-p[i]]) p[i]++, ans = max(ans, query(pos[i+p[i]], 2*p[i]));
                if(i+p[i]>mx) mx = i+p[i], id = i;
            }
            printf("%lld
    ", ans);
        }
    } sam;
    
    int main() {
        scanf("%s", s + 1);
        n = strlen(s + 1);
        for(int i = 1; i <= n; i++)
            sam.extend(s[i]-'a', i);
        sam.getSize(n);
        sam.solve();
        return 0;
    }
    
    /*
    */
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  • 原文地址:https://www.cnblogs.com/CJLHY/p/9824966.html
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