HDU 4513 吉哥系列故事完美队形II （manacher）-优快云博客

本文介绍了一种用于寻找字符串中最长回文子串的方法，并确保左侧部分为非递减序列。通过Manacher算法实现，适用于长度不超过10^5的输入串。代码示例使用C++编写，展示了如何构建特殊字符串并计算回文半径。

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题意:

$N<=10^5的串, 求最长回文串, 并且左边的部分是不降的$

分析:

$计算p数组的时候判断一下就好了, - - 注意'\#'$

代码:

//
//  Created by TaoSama on 2015-11-03
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 2e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, k, p[N];
int a[N], s[N];

int manacher() {
    s[0] = '@'; s[1] = '#'; n = 2;
    for(int i = 0; i < k; ++i)
        s[n++] = a[i], s[n++] = '#';
    s[n] = 0;

    int mx = 0, id, ret = 0;
    for(int i = 1; i < n; ++i) {
        p[i] = mx > i ? min(mx - i, p[2 * id - i]) : 1;
        while(s[i - p[i]] == s[i + p[i]]) {
            if(p[i] == 1 || s[i + p[i]] == '#') {}
            else {
                if(s[i + p[i]] > s[i + p[i] - 2]) break;
            }
            ++p[i];
        }
        if(mx < i + p[i]) mx = i + p[i], id = i;
        ret = max(ret, p[i] - 1);
    }
    return ret;
}

int main() {
#ifdef LOCAL
    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    int t; scanf("%d", &t);
    while(t--) {
        scanf("%d", &k);
        for(int i = 0; i < k; ++i) scanf("%d", a + i);
        printf("%d\n", manacher());
    }
    return 0;
}