HDU6599:求本质不同的子串(回文串)长度及数量

@(hdu6599:求本质不同的回文串长度及数量)
欢迎交流

hdu6599题意:

传送门: here

求有多少个回文串的前\(⌈ \frac {len}{2} ⌉\)个字符也是回文串。(两组解可重复)
将这些回文串按长度分类,分别输出长度为\(1,2,...,n\)的合法串的数量。

前期分析:
首先考虑回文串怎么求?\(manacher\)可以求出以\(i\)点为端点或中心的回文串数量。
但是求出来后你没法判断他的前半部分是否也是回文串。
推荐一个南京\(manacher\)练习题:here
还有注意一个字符串的回文串的数量太多了!
就算你能快速判断一个回文串是否满足条件,你不能依次判断每个回文串是否可行呀!

有一个结论:

一个字符串本质不同的回文串的数量级是\(O(n)\)的。

然后只要能快速求出一个子串且是回文串在原串的出现次数本题就解决了。
有个类似的题是求子串的,方法可以后缀数组也可以后缀自动机:here
一看数据范围\(|S|\le 3e5\),然后尝试我就挣扎了好多\(O(nlog(n))\)的做法,当然也有\(O(n)\)的做法(哭戚戚qwq

manacher+后缀自动机+倍增 \(O(nlog(n))\)

manacher得到本质不同的回文串

  • 首先所有本质不同的回文串都会在\(manacher\)扩展的时候遍历到,然后\(manacher\)是均摊\(O(n)\)的复杂度。
  • 那现在我可以大致获得\(O(n)\)级别的回文串,我再\(hash\)去重一下就可以得到所有本质不同的回文串,顺便\(hash\)判断它是否合法。

去重之类的可以\(hash(取模或uLL),map,unordered\_map\),自信不会\(hash\)冲突的话可以用一个\(bool\)数组代替\(map\)

统计子串在母串中出现次数

  • 这样接下来我只要能求出这个字符串\((l,r)\)在母串中出现的次数,本题就结束了。
  • 后缀自动机里:\(|endpos(u)|\)表示状态\(u\)这个集合内字符串出现的次数。
  • 我从字符串\((l,r)\)右端点字符所在的状态沿着后缀连接树网上跳,遇到一个状态的\(minlen(u)\)大于等于这个字符串的长度就停下来,这个子串就肯定在状态\(u\)集合里面,出现次数即\(|endpos(u)|\)
  • 暴力跳肯定不行,利用\(lca\)的倍增算法预处理一下就行了,你可以在后缀连接树上搜一遍,也可以拓扑排序后\(for\)预处理一遍,也可以基数排序后\(for\)一下预处理一遍。

对了,如果内存比较紧张,可以不用另开数组来倍增,直接用后缀自动机的\(nex\)数组,但是,一定要注意初始化!!!

Code

#pragma comment(linker, "/STACK:102400000,102400000")
#include<bits/stdc++.h>
#define fi first
#define se second
#define endl '\n'
#define o2(x) (x)*(x)
#define BASE_MAX 30
#define mk make_pair
#define eb emplace_back
#define all(x) (x).begin(), (x).end()
#define clr(a, b) memset((a),(b),sizeof((a)))
#define iis std::ios::sync_with_stdio(false); cin.tie(0)
#define my_unique(x) sort(all(x)),x.erase(unique(all(x)),x.end())
using namespace std;
#pragma optimize("-O3")
typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int, int> pii;
inline LL read() {
    LL x = 0;int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x = f ? -x : x;
}
inline void write(LL x, bool f) {
    if (x == 0) {putchar('0'); if(f)putchar('\n');else putchar(' ');return;}
    if (x < 0) {putchar('-');x = -x;}
    static char s[23];
    int l = 0;
    while (x != 0)s[l++] = x % 10 + 48, x /= 10;
    while (l)putchar(s[--l]);
    if(f)putchar('\n');else putchar(' ');
}
int lowbit(int x) { return x & (-x); }
template<class T>T big(const T &a1, const T &a2) { return a1 > a2 ? a1 : a2; }
template<typename T, typename ...R>T big(const T &f, const R &...r) { return big(f, big(r...)); }
template<class T>T sml(const T &a1, const T &a2) { return a1 < a2 ? a1 : a2; }
template<typename T, typename ...R>T sml(const T &f, const R &...r) { return sml(f, sml(r...)); }
void debug_out() { cerr << '\n'; }
template<typename T, typename ...R>void debug_out(const T &f, const R &...r) {cerr << f << " ";debug_out(r...);}
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]: ", debug_out(__VA_ARGS__);

#define print(x) write(x);

const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int HMOD[] = {1000000009, 1004535809};
const LL BASE[] = {1572872831, 1971536491};
const int mod = 998244353;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MXN = 3e5 + 7;
const uLL base = 19260817;
char s[MXN];
int p[MXN], bin[22];
LL ANS[MXN];
uLL hs[MXN], bs[MXN];
inline uLL get_hash(int l, int r) {
    return hs[r] - hs[l-1]*bs[r-l+1];
}
unordered_map<uLL, int> ump;
int n;
struct Suffix_Automaton {
    static const int maxn = 3e5 + 105;
    static const int MAXN = 3e5 + 5;
    //basic
//    map<char,int> nex[maxn * 2];
    int nex[maxn*2][26];
    int link[maxn * 2], len[maxn * 2];
    int last, cnt;
    LL tot_c;//不同串的个数
    //extension
    int cntA[MAXN * 2], A[MAXN * 2];//辅助拓扑更新
    int nums[MAXN * 2];//每个节点代表的所有串的出现次数
    int dep[MAXN*2], pos[MAXN*2];
    void clear() {
        tot_c = 0;
        last = cnt = 1;
        link[1] = len[1] = 0;
//        nex[1].clear();
        memset(nex[1], 0, sizeof(nex[1]));
        for(int i = 0; i <= 2 * n + 1; ++i) nums[i] = 0;
    }
    void add(int c, int id) {
        int p = last, np = ++ cnt;
//        nex[cnt].clear();
        memset(nex[cnt], 0, sizeof(nex[cnt]));
        len[np] = len[p] + 1;
        nums[np] = 1;
        pos[id] = np;
        last = np;
        while (p && !nex[p][c])nex[p][c] = np, p = link[p];
        if (!p)link[np] = 1, tot_c += len[np] - len[link[np]];
        else {
            int q = nex[p][c];
            if (len[q] == len[p] + 1)link[np] = q, tot_c += len[np] - len[link[np]];
            else {
                int nq = ++cnt;
                len[nq] = len[p] + 1;
//                nex[nq] = nex[q];
                memcpy(nex[nq], nex[q], sizeof(nex[q]));
                link[nq] = link[q];
                link[np] = link[q] = nq;
                tot_c += len[np] - len[link[np]];
                while (nex[p][c] == q)nex[p][c] = nq, p = link[p];
            }
        }
    }
    void build(int n) {
//        memset(cntA, 0, sizeof cntA), memset(nums, 0, sizeof nums);
        for(int i = 0; i <= cnt; ++i) cntA[i] = 0;
        for (int i = 1; i <= cnt; i++)cntA[len[i]]++;
        for (int i = 1; i <= n; i++)cntA[i] += cntA[i - 1];
        for (int i = cnt; i >= 1; i--)A[cntA[len[i]]--] = i;
        //更行主串节点
//        int temps = 1;
//        for (int i = 1; i <= n; ++i) {
//            nums[temps = nex[temps][s[i] - 'a']] = 1;
//            debug(temps, s[i-1])
//        }
        //拓扑更新
        for (int i = cnt, x; i >= 1; i--) {
            //basic
            x = A[i];
            nums[link[x]] += nums[x];
        }
//        for(int i = 1; i <= cnt; ++i) debug(nums[i])
        for (int i = 1, t; i <= cnt; i++) {
            t = A[i];
            dep[t] = dep[link[t]] + 1;
            nex[t][0] = link[t];
//            for (int j = 1; bin[j] <= dep[t]; j++)
            for (int j = 1; j < 20; j++)
                if(bin[j] <= dep[t]) nex[t][j] = nex[nex[t][j - 1]][j - 1];
                else nex[t][j] = 0;
        }
    }
//    void DEBUG() {
//        for (int i = cnt; i >= 1; i--) {
//            printf("nums[%d]=%d numt[%d]=%d len[%d]=%d link[%d]=%d\n", i, nums[i], i, nums[i], i, len[i], i, link[i]);
//        }
//    }
    int query(int l, int r) {
        int mid = pos[r];
        for (int i = 19; i >= 0; i--) {
            int t = nex[mid][i];
            if (len[t] >= r - l + 1) mid = t;
        }
//        ans += siz[mid];
        ANS[r-l+1] += nums[mid];
//        debug(l, r, nums[mid])
    }
} sam;

void manacher() {
    int mx = 0, id;
    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]]) {
            if(ump.find(get_hash(i - p[i], i + p[i] + 1)) == ump.end()) {
                ump[get_hash(i - p[i], i + p[i] + 1)] = 1;
//                debug(gethash(i - p[i], i + p[i] + 1), i - p[i], i + p[i] + 1)
                int l = i - p[i], r = i + p[i] + 1;
                if(get_hash(l, (l+r)/2) == get_hash((l+r)/2+1, r)) {
                    sam.query(i - p[i], i + p[i] + 1);
                }
            }
            p[i]++;
        }
        if (p[i] + i > mx)mx = p[i] + i, id = i;
    }
    mx = 0;
    for (int i = 1; i <= n; i++) {
        if (mx > i)p[i] = min(mx - i - 1, p[2 * id - i]);
        else {
            p[i] = 1;
            if(ump.find(get_hash(i - p[i] + 1, i + p[i] - 1)) == ump.end()) {
                ump[get_hash(i - p[i] + 1, i + p[i] - 1)] = 1;
//                debug(gethash(i - p[i] + 1, i + p[i] - 1), i - p[i] + 1, i + p[i] - 1)
                int l = i - p[i] + 1, r = i + p[i] - 1;
                if(get_hash(l, (l+r)/2) == get_hash((l+r)/2, r)) {
                    sam.query(i - p[i] + 1, i + p[i] - 1);
                }
            }
        }
        while (s[i + p[i]] == s[i - p[i]]) {
            if(ump.find(get_hash(i - p[i], i + p[i])) == ump.end()) {
                ump[get_hash(i - p[i], i + p[i])] = 1;
//                debug(gethash(i - p[i], i + p[i]), i - p[i], i + p[i])
                int l = i - p[i], r = i + p[i];
                if(get_hash(l, (l+r)/2) == get_hash((l+r)/2, r)) {
                    sam.query(i - p[i], i + p[i]);
                }
            }
            p[i]++;
        }
        if (p[i] + i > mx)mx = p[i] + i, id = i;
    }
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("/home/cwolf9/CLionProjects/ccc/in.txt", "r", stdin);
    //freopen("/home/cwolf9/CLionProjects/ccc/out.txt", "w", stdout);
#endif
//    int tim = read();
    bin[0] = 1; bs[0] = 1;
    for (int i = 1; i < 20; i++) bin[i] = bin[i - 1] << 1;
    for(int i = 1;  i<= 300005; ++i) bs[i] = bs[i-1] * base;
    while (~scanf("%s", s + 1)) {
//        debug(s+1)
        n = strlen(s + 1);
        for(int i = 1; i <= n; ++i) hs[i] = hs[i-1] * base + s[i];
        sam.clear();
        ump.clear();
        for(int i = 0; i <= n + 2; ++i) p[i] = 0;
        for(int i = 1; i <= n; ++i) sam.add(s[i] - 'a', i), ANS[i] = 0;
        sam.build(n);
        manacher();
        for (int i = 1; i <= n; ++i) if (i == n) write(ANS[i], true); else write(ANS[i], false);
    }
#ifndef ONLINE_JUDGE
    cout << "time cost:" << clock() << "ms" << endl;
#endif
    return 0;
}

manacher+后缀数组+二分 \(O(nlog(n))\)

manacher得到本质不同的回文串

  • 首先所有本质不同的回文串都会在\(manacher\)扩展的时候遍历到,然后\(manacher\)是均摊\(O(n)\)的复杂度。
  • 那现在我可以大致获得\(O(n)\)级别的回文串,我再\(hash\)去重一下就可以得到所有本质不同的回文串,顺便\(hash\)判断它是否合法。

去重之类的可以\(hash(取模或uLL),map,unordered\_map\),自信不会\(hash\)冲突的话可以用一个\(bool\)数组代替\(map\)

统计子串在母串中出现次数

  • 求两个后缀串的\(lcp\)就是直接\(height\)最小值,\(RMQ\)预处理一下就行。这是后缀数组的板子,大家都会。

  • 求一个子串出现次数,利用\(height\)数组,将排序后的后缀分组,把\(height\)大于等于\(len(l,r)\)的后缀分成一组,然后统计数量即可,这是国家集训队罗大佬论文里常规套路。
  • 对于分组后左右端点的确定,二分一下即可。(不过要注意细节

网上板子的\(RMQ\_Query(l,r)\)用着不习惯,自己改成自己习惯的写法,二分的时候注意一下细节即可。别忘了\(height[i]\)\(sa[i-1]\)\(sa[i]\)\(lcp\)

后缀数组写法可以参考一下这个:here

Code

#pragma comment(linker, "/STACK:102400000,102400000")
#include<bits/stdc++.h>
#define fi first
#define se second
#define endl '\n'
#define o2(x) (x)*(x)
#define BASE_MAX 30
#define mk make_pair
#define eb emplace_back
#define all(x) (x).begin(), (x).end()
#define clr(a, b) memset((a),(b),sizeof((a)))
#define iis std::ios::sync_with_stdio(false); cin.tie(0)
#define my_unique(x) sort(all(x)),x.erase(unique(all(x)),x.end())
using namespace std;
#pragma optimize("-O3")
typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int, int> pii;
inline LL read() {
    LL x = 0;int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x = f ? -x : x;
}
inline void write(LL x, bool f) {
    if (x == 0) {putchar('0'); if(f)putchar('\n');else putchar(' ');return;}
    if (x < 0) {putchar('-');x = -x;}
    static char s[23];
    int l = 0;
    while (x != 0)s[l++] = x % 10 + 48, x /= 10;
    while (l)putchar(s[--l]);
    if(f)putchar('\n');else putchar(' ');
}
int lowbit(int x) { return x & (-x); }
template<class T>T big(const T &a1, const T &a2) { return a1 > a2 ? a1 : a2; }
template<typename T, typename ...R>T big(const T &f, const R &...r) { return big(f, big(r...)); }
template<class T>T sml(const T &a1, const T &a2) { return a1 < a2 ? a1 : a2; }
template<typename T, typename ...R>T sml(const T &f, const R &...r) { return sml(f, sml(r...)); }
void debug_out() { cerr << '\n'; }
template<typename T, typename ...R>void debug_out(const T &f, const R &...r) {cerr << f << " ";debug_out(r...);}
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]: ", debug_out(__VA_ARGS__);

#define print(x) write(x);

const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int HMOD[] = {1000000009, 1004535809};
const LL BASE[] = {1572872831, 1971536491};
const int mod = 1e7 + 7;
const int MOD = 1e7 + 7;//998244353
const int INF = 0x3f3f3f3f;
const int MXN = 3e5 + 7;
const uLL base = 19260817;
char s[MXN];
int p[MXN], bin[22], lg2[MXN];
LL ANS[MXN];
uLL hs[MXN], bs[MXN];
unordered_map<uLL,bool> ump;
inline uLL get_hash(int l, int r) {
    return ((hs[r] - hs[l-1]*bs[r-l+1]));
}
int n;
//后缀数组(SA[i]存放排名第i大的后缀首字符的下标)
//名次数组(rank[i]存放Suffix(i)的优先级(名次))
//height数组:height[i]是Suffix(sa[i-1])和Suffix(sa[i])的最长公共前缀长度
//SA,R,H的下标都是 0~n 其中多包括了一个空字符串
struct Suffix_Array {
    static const int N = 3e5 + 7;
    int n, len, s[N], M;
    int sa[N], rnk[N], height[N];
    int tmp_one[N], tmp_two[N], c[N];
    int dp[N][21];
    void init_str(char *str, int _n) {
        len = _n;
        n = len + 1;
        for (int i = 0; i < len; ++i) s[i] = str[i];
        s[len] = '\0';
    }
    void build_sa(int m = 128);
    void calc_height(int n);
    void Out(char *str);
    void query(int l, int r);
    void RMQ_init(int n);
    int RMQ_query(int l, int r);
}sam;
void Suffix_Array::Out(char *str) {
    puts ("/*Suffix*/");
    for (int i=0; i<n; ++i) {
        printf ("%s\n", str+sa[i]);
    }
}
int Suffix_Array::RMQ_query(int l, int r) {//看自己需求自由变换
    int k = lg2[r - l + 1];
//    int k = 0; while (1<<(k+1) <= r - l + 1) k++;
    return min(dp[l][k], dp[r-(1<<k)+1][k]);
}
void Suffix_Array::RMQ_init(int n) {
    for (int i=0; i<n; ++i) dp[i][0] = height[i];
    for (int j=1; (1<<j)<=n; ++j) {
        for (int i=0; i+(1<<j)-1<n; ++i) {
            dp[i][j] = std::min (dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
        }
    }
}
void Suffix_Array::calc_height(int n) {
    for (int i=0; i<=n; ++i) rnk[sa[i]] = i;
    int k = height[0] = 0;
    for (int i=0; i<n; ++i) {
        if (k) k--;
        int j = sa[rnk[i]-1];
        while (s[i+k] == s[j+k]) k++;
        height[rnk[i]] = k;
    }
}
//m = max(r[i]) + 1,一般字符128足够了
void Suffix_Array::build_sa(int m) {
    int i, j, p, *x = tmp_one, *y = tmp_two;
    for (i=0; i<m; ++i) c[i] = 0;
    for (i=0; i<n; ++i) c[x[i]=s[i]]++;//此时第一关键字是x[i],第二关键字是i
    for (i=1; i<m; ++i) c[i] += c[i-1];
    for (i=n-1; i>=0; --i) sa[--c[x[i]]] = i;//排第几的后缀是i
    for (j=1; j<=n; j<<=1) {//y就是第二关键字从小到大的位置
        //y[i]表示第二关键字排名为i的数,第一关键字的位置
        for (p=0, i=n-j; i<n; ++i) y[p++] = i;//这些数的第二关键字为0
        for (i=0; i<n; ++i) if (sa[i] >= j) y[p++] = sa[i] - j;//按rank顺序,1<<(j+1)的第二半的rank。sa[i]把自己交给了sa[i]-j
        //现在第二关键字已经有序,在此基础上按第一关键字排序
        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];//排第几的后缀是y[i]
        std::swap (x, y);
        for (p=1, x[sa[0]]=0, i=1; i<n; ++i) {//排完序后更新第一关键字
            x[sa[i]] = (y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+j] == y[sa[i]+j] ? p - 1 : p++);
        }
        if(p >= n) break;
        m=p;
    }
    calc_height(n-1);
    RMQ_init(n);
}
void Suffix_Array::query(int l, int r) {
    -- l, -- r;
    int len = r - l + 1, L = 1, R = rnk[l]-1, mid, ans = -1;
    while(L <= R) {
        mid = (L + R) >> 1;
        if(RMQ_query(mid+1,  rnk[l]) >= len) ans = mid, R = mid - 1;
        else L = mid + 1;
    }
    int tmp = ans;
    L =  rnk[l] + 1, R = n - 1, ans = -1;
    while(L <= R) {
        mid = (L + R) >> 1;
        if(RMQ_query( rnk[l] + 1, mid) >= len) ans = mid, L = mid + 1;
        else R = mid - 1;
    }
    if(ans == -1 && tmp == -1) ANS[r-l+1] ++;
    else if(ans == -1) ANS[r-l+1] +=  rnk[l] - tmp + 1;
    else if(tmp == -1) ANS[r-l+1] += ans - rnk[l] + 1;
    else ANS[r-l+1] += ans - tmp + 1;
//    printf("%d %d %c %d %d %d\n", l, r, s[l],  rnk[l], ans, tmp);
}
void manacher() {
    int mx = 0, id;
    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]]) {
            if(ump.find(get_hash(i - p[i], i + p[i] + 1)) == ump.end()) {
                ump[get_hash(i - p[i], i + p[i] + 1)] = true;
                int l = i - p[i], r = i + p[i] + 1;
                if(get_hash(l, (l+r)/2) == get_hash((l+r)/2+1, r)) {
                    sam.query(i - p[i], i + p[i] + 1);
                }
            }
            p[i]++;
        }
        if (p[i] + i > mx)mx = p[i] + i, id = i;
    }
    mx = 0;
    for (int i = 1; i <= n; i++) {
        if (mx > i)p[i] = min(mx - i - 1, p[2 * id - i]);
        else {
            p[i] = 1;
            if(ump.find(get_hash(i - p[i] + 1, i + p[i] - 1)) == ump.end()) {
                ump[get_hash(i - p[i] + 1, i + p[i] - 1)] = true;
                int l = i - p[i] + 1, r = i + p[i] - 1;
                if(get_hash(l, (l+r)/2) == get_hash((l+r)/2, r)) {
                    sam.query(i - p[i] + 1, i + p[i] - 1);
                }
            }
        }
        while (s[i + p[i]] == s[i - p[i]]) {
            if(ump.find(get_hash(i - p[i], i + p[i])) == ump.end()) {
                ump[get_hash(i - p[i], i + p[i])] = true;
                int l = i - p[i], r = i + p[i];
                if(get_hash(l, (l+r)/2) == get_hash((l+r)/2, r)) {
                    sam.query(i - p[i], i + p[i]);
                }
            }
            p[i]++;
        }
        if (p[i] + i > mx)mx = p[i] + i, id = i;
    }
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("/home/cwolf9/CLionProjects/ccc/in.txt", "r", stdin);
    //freopen("/home/cwolf9/CLionProjects/ccc/out.txt", "w", stdout);
#endif
//    int tim = read();
    bin[0] = 1;
    for (int i = 1; i < 22; i++) bin[i] = bin[i - 1] << 1;
    for (int i = 2; i < MXN; ++i) lg2[i] = lg2[i >> 1] + 1;
    bs[0] = 1;
    for(int i = 1;  i< MXN; ++i) bs[i] = (bs[i-1] * base);
    while (~scanf("%s", s + 1)) {
        n = strlen(s + 1);
        sam.init_str(s + 1, n);
        sam.build_sa();
//        for(int i = 0; i <= n; ++i) printf("%d ", sam.sa[i]); printf("\n");
        for(int i = 1; i <= n; ++i) hs[i] = (hs[i-1] * base + s[i]);
        ump.clear();
        for(int i = 0; i <= n + 2; ++i) p[i] = 0;
        for(int i = 1; i <= n; ++i) ANS[i] = 0;
        manacher();
        for (int i = 1; i <= n; ++i) if (i == n) write(ANS[i], true); else write(ANS[i], false);
    }
#ifndef ONLINE_JUDGE
    cout << "time cost:" << clock() << "ms" << endl;
#endif
    return 0;
}

回文树(回文自动机) \(O(n)\)

回文树可以干啥?

假设我们有一个串\(S\)\(S\)下标从\(0\)开始,则回文树能做到如下几点:

1.求串\(S\)前缀\(0 - i\)本质不同回文串的个数(两个串长度不同或者长度相同且至少有一个字符不同便是本质不同)
2.求串\(S\)每一个本质不同回文串出现的次数
3.求串\(S\)回文串的个数(其实就是1和2结合起来)
4.求以下标\(i\)结尾的回文串的个数

\(what??\)前面费劲力气干的事情,回文树\(O(n)\)全给你搞定。
本题我从这个博客找了板子,就是一个\(pos[]\)数组记录原字符串端点对应回文树上的端点,再用\(hash\)判断字符串是否相等。随便改改就\(ac\)了!??

简直自闭,出一个模板题!!醉了。。

code

#pragma comment(linker, "/STACK:102400000,102400000")
#include<bits/stdc++.h>
#define fi first
#define se second
#define endl '\n'
#define o2(x) (x)*(x)
#define BASE_MAX 30
#define mk make_pair
#define eb emplace_back
#define all(x) (x).begin(), (x).end()
#define clr(a, b) memset((a),(b),sizeof((a)))
#define iis std::ios::sync_with_stdio(false); cin.tie(0)
#define my_unique(x) sort(all(x)),x.erase(unique(all(x)),x.end())
using namespace std;
#pragma optimize("-O3")
typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int, int> pii;
inline LL read() {
    LL x = 0;int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
    return x = f ? -x : x;
}
inline void write(LL x, bool f) {
    if (x == 0) {putchar('0'); if(f)putchar('\n');else putchar(' ');return;}
    if (x < 0) {putchar('-');x = -x;}
    static char s[23];
    int l = 0;
    while (x != 0)s[l++] = x % 10 + 48, x /= 10;
    while (l)putchar(s[--l]);
    if(f)putchar('\n');else putchar(' ');
}
int lowbit(int x) { return x & (-x); }
template<class T>T big(const T &a1, const T &a2) { return a1 > a2 ? a1 : a2; }
template<typename T, typename ...R>T big(const T &f, const R &...r) { return big(f, big(r...)); }
template<class T>T sml(const T &a1, const T &a2) { return a1 < a2 ? a1 : a2; }
template<typename T, typename ...R>T sml(const T &f, const R &...r) { return sml(f, sml(r...)); }
void debug_out() { cerr << '\n'; }
template<typename T, typename ...R>void debug_out(const T &f, const R &...r) {cerr << f << " ";debug_out(r...);}
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]: ", debug_out(__VA_ARGS__);

#define print(x) write(x);

const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int HMOD[] = {1000000009, 1004535809};
const LL BASE[] = {1572872831, 1971536491};
const int mod = 1e7 + 7;
const int MOD = 1e7 + 7;//998244353
const int INF = 0x3f3f3f3f;
const int MXN = 3e5 + 7;
const uLL base = 19260817;
char s[MXN];
int n;
LL ANS[MXN];
uLL hs[MXN], bs[MXN];
inline uLL get_hash(int l, int r) {
    return hs[r] - hs[l-1]*bs[r-l+1];
}
struct Palindromic_Tree {
    static const int MAXN = 600005 ;
    static const int CHAR_N = 26 ;
    int next[MAXN][CHAR_N];//next指针,next指针和字典树类似,指向的串为当前串两端加上同一个字符构成
    int fail[MAXN];//fail指针,失配后跳转到fail指针指向的节点
    int cnt[MAXN];
    int num[MAXN];
    int len[MAXN];//len[i]表示节点i表示的回文串的长度
    int S[MAXN];//存放添加的字符
    int last;//指向上一个字符所在的节点,方便下一次add
    int n;//字符数组指针
    int p;//节点指针
    int pos[MAXN];
    int newnode(int l) {//新建节点
        for (int i = 0; i < CHAR_N; ++i) next[p][i] = 0;
        cnt[p] = 0;
        num[p] = 0;
        len[p] = l;
        return p++;
    }
    void init() {//初始化
        p = 0;
        newnode(0);
        newnode(-1);
        last = 0;
        n = 0;
        S[n] = -1;//开头放一个字符集中没有的字符,减少特判
        fail[0] = 1;
    }
    int get_fail(int x) {//和KMP一样,失配后找一个尽量最长的
        while (S[n - len[x] - 1] != S[n]) x = fail[x];
        return x;
    }
    void add(int c, int id) {
        c -= 'a';
        S[++n] = c;
        int cur = get_fail(last);//通过上一个回文串找这个回文串的匹配位置
        if (!next[cur][c]) {//如果这个回文串没有出现过,说明出现了一个新的本质不同的回文串
            int now = newnode(len[cur] + 2);//新建节点
            fail[now] = next[get_fail(fail[cur])][c];//和AC自动机一样建立fail指针,以便失配后跳转
            next[cur][c] = now;
            num[now] = num[fail[now]] + 1;
        }
        last = next[cur][c];
        cnt[last] ++;
        pos[last] = id;
    }
    void count() {
        for (int i = p - 1; i >= 0; --i) cnt[fail[i]] += cnt[i];
        //父亲累加儿子的cnt,因为如果fail[v]=u,则u一定是v的子回文串!
        for(int i = 0, tmp; i < p; ++i) {
            tmp = pos[i];
            if(len[i] % 2 == 0 && get_hash(tmp - len[i] + 1, tmp - len[i]/2) == get_hash(tmp - len[i]/2 + 1, tmp)) ANS[len[i]] += cnt[i];
            else if((len[i] & 1) && get_hash(tmp - len[i] + 1, tmp - len[i]/2) == get_hash(tmp - len[i]/2, tmp)) ANS[len[i]] += cnt[i];
        }
    }
} pt;
int main() {
#ifndef ONLINE_JUDGE
    freopen("/home/cwolf9/CLionProjects/ccc/in.txt", "r", stdin);
    //freopen("/home/cwolf9/CLionProjects/ccc/out.txt", "w", stdout);
#endif
//    int tim = read();
    bs[0] = 1;
    for(int i = 1;  i< MXN; ++i) bs[i] = (bs[i-1] * base);
    while (~scanf("%s", s + 1)) {
        n = strlen(s + 1);
        for (int i = 1; i <= n; ++i) ANS[i] = 0, hs[i] = hs[i-1] * base + s[i];
        pt.init();
        for(int i = 1; i <= n; ++i) pt.add(s[i], i);
        pt.count();
        for (int i = 1; i <= n; ++i) if (i == n) write(ANS[i], true); else write(ANS[i], false);
    }
#ifndef ONLINE_JUDGE
    cout << "time cost:" << clock() << "ms" << endl;
#endif
    return 0;
}
posted @ 2019-07-26 21:48 Cwolf9 阅读(...) 评论(...) 编辑 收藏

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