ZZULI 1876: 蛤玮的项链 Hash + 二分

蛤玮向心仪的妹子送了一条项链,这条项链是由小写字母构成的首尾相接的字符串,妹子看了看项链对蛤玮说,"我希望它是对称的",蛤玮想了想之后决定,从项链上截取出一段,这段如果是回文的话那么妹子戴起来就是对称的了.由于蛤玮会魔法,他可以把项链上的某一个字母变成任意另一个字母,但由于魔力限制他最多只能变两次,现在蛤玮想知道他能截取出的项链的最长长度是多少.为了简单,我们假设蛤玮截取出的长度必须是奇数.

样例串改变一个字母变成abcbaaa,整个项链便可转成回文aabcbaa.

思路：（dzs教我的）。由于是循环的，那么将s变为ss，类似用hash求以i为中心的最长回文的长度，对于每一个位置i，先二分到pos1，那么pos1-i-（i-pos1+i）为当前的回文段，pos1-=2，相当于修改一次操作，继续二分到一个位置pos2.如此做两次，就相当于两次修改操作

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
#include <map>
#include <string>
using namespace std;
const int x = 123;
const int N = 200005;
unsigned long long H1[N], H2[N], xp[N];
char s[N];
int n, m;
void initHash() {
    H1[n] = H2[n] = 0;
    int t = 0;
    for(int i = n - 1; i >= 0; --i) {
        H2[i] = H2[i + 1] * x + s[i];
        H1[i] = H1[i + 1] * x + s[t++];
    }
    xp[0] = 1;
    for(int i = 1; i <= n; ++i) xp[i] = xp[i - 1] * x;
}
unsigned long long getHash(int i, int L, int f) {
    unsigned long long h;
    if(f == 1)
        h = H1[i] - H1[i + L] * xp[L];
    else
        h = H2[i] - H2[i + L] * xp[L];
    return h;
}
void init() {
    scanf("%s", s);
    m = strlen(s);
    for(int i = 0; i < m; ++i) s[i + m] = s[i];
    n = m << 1;
    initHash();
}
int get(int i) {
    int L = 0, R = i + 1;
    while(R - L > 1) {
        int M = (L + R) >> 1;
        if(n - i + M <= n && i + 1 + M <= n && getHash(n - i, M, 1) == getHash(i + 1, M, 2))
         L = M;
        else R = M;
    }
    return L;
}
int change(int i, int cen) {
    int L = 0, R = i + 2;
    while(R - L > 1) {
        int M = (L + R) >> 1;
        if(n - i - 1 + M <= n && 2 * cen - i + M <= n && getHash(n - i - 1, M, 1) == getHash(2 * cen - i, M, 2))
        L = M;
        else R = M;
    }
    return L;
}
int solve() {
    int pos1, pos2, pos3, ls1, ls2;
    if(m <= 5) return m;
    int ans = 5;
    for(int i = 3; i < n; ++i)
    {
        int x = get(i);
        pos1 = i - x;
        if(x + 2 + i < n) pos1 -= 2;
        ls1 = change(pos1, i);
        pos2 = pos1 - ls1 + 1;
        if(pos2 == 1 && i - pos2 + i + 1 < n) pos3 = 0;
        else if(pos2 == 0) pos3 = pos2;
        else {
            pos3 = pos2;
            if(i - pos2 + i + 2 < n) {
                pos2 -= 2;
                ls2 = change(pos2, i);
                pos3 = pos2 - ls2 + 1;
            }
        }
        ans = max(ans, (i - pos3) * 2 + 1);
    }
    return min(m, ans);
}
int main() {
 // freopen("in", "r", stdin);
    int _; scanf("%d", &_);
    while(_ --) {
        init();
        int ans = solve();
        if(ans % 2 == 0) ans--;
        printf("%d
", ans);
    }
    return 0;
}

/**************************************************************
    Problem: 1876
    User: atrp
    Language: C++
    Result: Accepted
    Time:2676 ms
    Memory:6208 kb
****************************************************************/

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ZZULI 1876: 蛤玮的项链 Hash + 二分

Description

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HINT