ZZULI 1876: 蛤玮的项链 Hash + 二分

Time Limit: 6 Sec Memory Limit: 128 MB
Submit: 153 Solved: 11

SubmitStatusWeb Board

Description

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.

Input

第一行整数T(1<=T<=10),表示数据组数.

每组数据一个字符串s,表示项链,|s|<=100000.

Output

每组数据输出一个数,最长的截取长度.

Sample Input

abcdaaa

Sample Output

HINT

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];elseh = 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\n", ans);}return 0;
}/**************************************************************Problem: 1876User: atrpLanguage: C++Result: AcceptedTime:2676 msMemory:6208 kb
****************************************************************/

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转载于:https://www.cnblogs.com/orchidzjl/p/5483554.html