【Codeforces 245H】Queries for Number of Palindromes（回文树）

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本文介绍了一道关于回文串查询的编程竞赛题目及其解决方案。通过对给定字符串进行预处理，实现快速计算任意区间内回文串的数量。

Description

给定一个长度不大于 $5e3$ 的串 $s$ ，每次询问 $s_{l,r}$ 有多少个子串是回文串。

Solution

发现串比较短，先预处理出所有的答案，直接回答即可。

Code

/************************************************
 * Au: Hany01
 * Date: Jun 22nd, 2018
 * Prob: [CF245H] Queries for Number of Palindromes
 * Email: hany01@foxmail.com
 * Institute: Yali High School
************************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read()
{
    register int _, __; register char c_;
    for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
    for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 5005;

LL Ans[maxn][maxn];
char s[maxn], ss[maxn];
int n, nn, len[maxn], sz[maxn], ch[maxn][26], v[maxn], beg[maxn], nex[maxn], e, las, tot, pos[maxn], fa[maxn], dep[maxn];

inline void add(int uu, int vv) { v[++ e] = vv, nex[e] = beg[uu], beg[uu] = e; }

inline void extend(int i)
{
    int p = las, c = ss[i] - 97;
    while (ss[i] != ss[i - len[p] - 1]) p = fa[p];
    if (!ch[p][c]) {
        int np = ++ tot, k = fa[p];
        while (ss[i] != ss[i - len[k] - 1]) k = fa[k];
        len[np] = len[p] + 2, fa[np] = ch[k][c], dep[np] = dep[ch[k][c]] + 1, ch[p][c] = np;
    }
    ++ sz[las = ch[p][c]];
}

void dfs(int u) {
    for (register int i = beg[u]; i; i = nex[i]) sz[v[i]] += sz[u], dfs(v[i]);
}

int main()
{
#ifdef hany01
    File("cf245h");
#endif

    scanf("%s", s + 1), n = strlen(s + 1), fa[1] = fa[0] = 1, len[1] = -1;
    For(i, 1, n) {
        tot = 1, las = 0, Set(ch, 0), e = 0, Set(beg, 0), Set(sz, 0);
        For(j, i, n) ss[j - i + 1] = s[j];
        nn = n - i + 1, add(1, 0);
        For(j, 1, nn) extend(j), Ans[i][i + j - 1] = Ans[i][i + j - 2] + dep[las];
    }

    for (static int m = read(), l, r; m --; )
        l = read(), r = read(), printf("%lld\n", Ans[l][r]);

    return 0;
}
//笑绿鬟邻女，倚窗犹唱，夕阳西下。
//    -- 蒋捷《女冠子·元夕》