Codeforces Gym 100548G The Problem to Slow Down You (Palindromic Tree 或 Hash水过) 2014西安现场赛G题

题目大意：

就是现在给你两个长度不超过20W的字符串, 都只包含小写字母, 求相同的回文串对数 (S, T), 其中S == T, S来自第一个字符串, T来自第二个字符串, S和T都是回文串

大致思路：

首先很容易想到的是Manacher + 后缀数组二分 + Hash的做法， 复杂度O(nlogn)， 可惜的是这个题目Hash容易被卡

试了好几次Hash之后试了一发二次Hash居然神奇地过了= =...不过这题用Hash还是太靠RP了, 而且幸好时间放的宽到了20s

另外一个正确的解法是用回文树Palindromic Tree, 对于串A, B分别建立回文树, 然后从根节点(奇数的根节点和偶数根节点分别进行一次) dfs, 如果有相同的回文串必然存在相同的代表结点, 其根节点到那个回文的结点的路径也是相同的

于是两棵树同时遍历, 因为Palindromic Tree可以在O(n)的时间复杂度中处理出每种回文串的个数, 所以计数就很方便了, 另外由于回文个数在O(n)以内, dfs复杂度也是O(n)

所以在O(n)的复杂度解出这题就很容易了

两种方法的代码如下：

Manacher + 后缀数组二分 + Hash的做法 ( RP因素较高 ) ：

( Hash大法好, Hash出奇迹(>_<) )

Result  :  Accepted     Memory  :  66000 KB     Time  :  9984 ms

/* * Author: Gatevin * Created Time:  2015/3/31 14:25:53 * File Name: Rin_Tohsaka.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;typedef unsigned long long ulint;#define maxn 200010char A[maxn], B[maxn], s[maxn << 1];int R[maxn << 1];const ulint mod = 1e9 + 10071;set<pair<ulint, ulint> > S;ulint H[maxn], xp[maxn];ulint H2[maxn], xp2[maxn];const ulint seed = 300007uLL;const ulint seed2 = 500009uLL;void initHash(char *s, int n){    H[0] = (ulint)(s[0] - 'a' + 1);    for(int i = 1; i < n; i++)        H[i] = (H[i - 1]*seed % mod + (ulint)(s[i] -'a' + 1)) % mod;    return;}void initHash2(char *s, int n){    H2[0] = (ulint)(s[0] - 'a' + 1);    for(int i = 1; i < n; i++)        H2[i] = H2[i - 1]*seed2 + (ulint)(s[i] - 'a' + 1);    return;}ulint askHash(int l, int r){    if(l == 0)        return H[r];    else        return (H[r] - H[l - 1]*xp[r - l + 1] % mod + mod) % mod;}ulint askHash2(int l, int r){    if(l == 0)        return H2[r];    else        return H2[r] - H2[l - 1]*xp2[r - l + 1];}int wa[maxn], wb[maxn], wv[maxn], Ws[maxn];int cmp(int *r, int a, int b, int l){    return r[a] == r[b] && r[a + l] == r[b + l];}void da(int *r, int *sa, int n, int m){    int *x = wa, *y = wb, *t, i, j, p;    for(i = 0; i < m; i++) Ws[i] = 0;    for(i = 0; i < n; i++) Ws[x[i] = r[i]]++;    for(i = 1; i < m; i++) Ws[i] += Ws[i - 1];    for(i = n - 1; i >= 0; i--) sa[--Ws[x[i]]] = i;    for(j = 1, p = 1; p < n; j <<= 1, m = p)    {        for(p = 0, i = n - j; i < n; i++) y[p++] = i;        for(i = 0; i < n; i++) if(sa[i] >= j) y[p++] = sa[i] - j;        for(i = 0; i < n; i++) wv[i] = x[y[i]];        for(i = 0; i < m; i++) Ws[i] = 0;        for(i = 0; i < n; i++) Ws[wv[i]]++;        for(i = 1; i < m; i++) Ws[i] += Ws[i - 1];        for(i = n - 1; i >= 0; i--) sa[--Ws[wv[i]]] = y[i];        for(t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; i++)            x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;    }    return;}int rank[maxn], height[maxn];void calheight(int *r, int *sa, int n){    int i, j, k = 0;    for(i = 1; i <= n; i++) rank[sa[i]] = i;    for(i = 0; i < n; height[rank[i++]] = k)        for(k ? k-- : 0, j = sa[rank[i] - 1]; r[i + k] == r[j + k]; k++);    return;}int dp[maxn][20];void initRMQ(int n){    for(int i = 1; i <= n; i++) dp[i][0] = height[i];    for(int j = 1; (1 << j) <= n; j++)        for(int i = 1; i + (1 << j) - 1 <= n; i++)            dp[i][j] = min(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);    return;}int askRMQ(int a, int b){    //int ra = rank[a], rb = rank[b];    int ra = a, rb = b;    if(ra > rb) swap(ra, rb);    int k = 0;    while((1 << (k + 1)) <= rb - ra + 1) k++;    return min(dp[ra][k], dp[rb - (1 << k) + 1][k]);}int calCnt(int l, int r, int n){    int rl = rank[l];    int lmost = rl, rmost = rl;    int L = rl + 1, R = n, mid;    while(L <= R)    {        mid = (L + R) >> 1;        if(askRMQ(rl + 1, mid) >= r - l + 1)        {            L = mid + 1;            rmost = mid;        }        else            R = mid - 1;    }    L = 1, R = rl - 1;    while(L <= R)    {        mid = (L + R) >> 1;        if(askRMQ(mid + 1, rl) >= r - l + 1)        {            R = mid - 1;            lmost = mid;        }        else            L = mid + 1;    }    return rmost - lmost + 1;}vector <pair<int, int> > pal;void Manacher(char *s, int *R, int n){    int mx = 0, p = 0;    R[0] = 1;    S.clear(), pal.clear();    for(int i = 1; i < n; i++)    {        if(mx > i) R[i] = min(R[2*p - i], mx - i);        else R[i] = 1;        while(s[i - R[i]] == s[i + R[i]])            R[i]++;        if(i + R[i] > mx)        {            for(int j = mx; j < i + R[i]; j++)            {                int l = 2*i - j, r = j;                l >>= 1;                r = (r & 1) ? r >> 1 : (r >> 1) - 1;                if(l > r) continue;                ulint hashvalue1 = askHash(l, r);                ulint hashvalue2 = askHash2(l, r);                set<pair<ulint, ulint> > :: iterator it = S.find(make_pair(hashvalue1, hashvalue2));                if(it == S.end())                {                    S.insert(make_pair(hashvalue1, hashvalue2));                    pal.push_back(make_pair(l, r));                }            }            mx = i + R[i], p = i;        }    }    return;}map<pair<ulint, ulint> , int> Ma, Mb;int ss[maxn], sa[maxn];int main(){    xp[0] = 1uLL;    xp2[0] = 1uLL;    for(int i = 1; i < maxn; i++)        xp[i] = xp[i - 1]*seed % mod, xp2[i] = xp2[i - 1]*seed2;    int T;    scanf("%d", &T);    for(int cas = 1; cas <= T; cas++)    {        scanf("%s", A);        scanf("%s", B);        int la = strlen(A), lb = strlen(B);                initHash(A, la);        initHash2(A, la);        s[0] = '@';        for(int i = 0; i < la; i++)            s[2*i + 1] = A[i], s[2*i + 2] = '#', ss[i] = A[i] - 'a' + 1;        s[2*la] = '$';        ss[la] = 0;        Manacher(s, R, 2*la);        da(ss, sa, la + 1, 280);        calheight(ss, sa, la);        initRMQ(la);        Ma.clear();        for(unsigned int i = 0, sz = pal.size(); i < sz; i++)            Ma[make_pair(askHash(pal[i].first, pal[i].second), askHash2(pal[i].first, pal[i].second))] = calCnt(pal[i].first, pal[i].second, la);                initHash(B, lb);        initHash2(B, lb);        s[0] = '@';        for(int i = 0; i < lb; i++)            s[2*i + 1] = B[i], s[2*i + 2] = '#', ss[i] = B[i] -'a' + 1;        s[2*lb] = '$';        ss[lb] = 0;        Manacher(s, R, 2*lb);        da(ss, sa, lb + 1, 28);        calheight(ss, sa, lb);        initRMQ(lb);        Mb.clear();        for(unsigned int i = 0, sz = pal.size(); i < sz; i++)            Mb[make_pair(askHash(pal[i].first, pal[i].second), askHash2(pal[i].first, pal[i].second))] = calCnt(pal[i].first, pal[i].second, lb);                lint ans = 0;        for(map<pair<ulint, ulint> , int> :: iterator it = Ma.begin(); it != Ma.end(); it++)            if(Mb[(*it).first] != 0)                ans += (lint)(*it).second*(lint)Mb[(*it).first];        printf("Case #%d: %I64d\n", cas, ans);    }    return 0;}

Palindromic Tree的做法：

Result  :  Accepted     Memory  :  53608 KB     Time  :  327 ms

/* * Author: Gatevin * Created Time:  2015/3/31 17:08:38 * File Name: Rin_Tohsaka.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;#define maxn 200010struct Palindromic_Tree{    struct node    {        int next[26];        int len;        int sufflink;        int times;//记录这个node代表的回文串出现的次数    };    node tree[maxn];    int L, len, suff;    char s[maxn];    void newnode()    {        L++;        for(int i = 0; i < 26; i++)            tree[L].next[i] = -1;        tree[L].len = tree[L].sufflink = tree[L].times = 0;        return;    }    void init()    {        L = 0, suff = 2;        newnode(), newnode();        tree[1].len = -1; tree[1].sufflink = 1;        tree[2].len = 0; tree[2].sufflink = 1;        return;    }    bool addLetter(int pos)    {        int cur = suff, curlen = 0;        int alp = s[pos] - 'a';        while(1)        {            curlen = tree[cur].len;            if(pos - 1 - curlen >= 0 && s[pos - 1 - curlen] == s[pos])                break;            cur = tree[cur].sufflink;        }        if(tree[cur].next[alp] != -1)        {            suff = tree[cur].next[alp];            tree[suff].times++;            return false;        }        newnode();        suff = L;        tree[L].len = tree[cur].len + 2;        tree[cur].next[alp] = L;        if(tree[L].len == 1)        {            tree[L].sufflink = 2;            tree[L].times++;            return true;        }        while(1)        {            cur = tree[cur].sufflink;            curlen = tree[cur].len;            if(pos - 1 - curlen >= 0 && s[pos - 1 - curlen] == s[pos])            {                tree[L].sufflink = tree[cur].next[alp];                break;            }        }        tree[L].times++;        return true;    }    void count()    {        for(int i = L; i > 0; i--)            tree[tree[i].sufflink].times += tree[i].times;        return;    }    void build()    {        init();        scanf("%s", s);        int length = strlen(s);        for(int i = 0; i < length; i++)            addLetter(i);        count();        return;    }};Palindromic_Tree A, B;/* * dfs从两份树的奇偶根节点开始向下, 有相同的回文串就加上数量乘积 * 只有有相同的才继续向下找 */lint dfs(int nowA, int nowB){    lint ret = 0;    for(int i = 0; i < 26; i++)        if(A.tree[nowA].next[i] != -1 && B.tree[nowB].next[i] != -1)            ret += (lint)A.tree[A.tree[nowA].next[i]].times * (lint)B.tree[B.tree[nowB].next[i]].times                + dfs(A.tree[nowA].next[i], B.tree[nowB].next[i]);    return ret;}int main(){    int T;    scanf("%d", &T);    for(int cas = 1; cas <= T; cas++)    {        A.build();        B.build();        lint ans = dfs(1, 1) + dfs(2, 2);        printf("Case #%d: %I64d\n", cas, ans);    }    return 0;}