BZOJ 2565 最长双回文串 (Manacher)

2565: 最长双回文串

Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 2658 Solved: 1346
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Description

顺序和逆序读起来完全一样的串叫做回文串。比如acbca是回文串，而abc不是（abc的顺序为“abc”，逆序为“cba”，不相同）。
输入长度为n的串S，求S的最长双回文子串T,即可将T分为两部分X，Y，（|X|,|Y|≥1）且X和Y都是回文串。

Input

一行由小写英文字母组成的字符串S。

Output

一行一个整数，表示最长双回文子串的长度。

Sample Input

baacaabbacabb

Sample Output

HINT

样例说明

从第二个字符开始的字符串aacaabbacabb可分为aacaa与bbacabb两部分，且两者都是回文串。

对于100%的数据，2≤|S|≤10^5

2015.4.25新加数据一组

Source

2012国家集训队Round 1 day2

析：首先先用Manacher算法处理出来，然后再求以 i 为开头和结尾的的最长回文串长度lx[i]，rx[i]，然后再求lx[i] + rx[i] 最大值即可。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-4;
const int maxn = 1e5 + 10;
const int maxm = 2e5 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

char s[maxn];
char ma[maxn<<1];
int len[maxn<<1];
int lx[maxn<<1], rx[maxn<<1];

void manacher(int n){
  int l = 0;
  ma[l++] = '$';
  ma[l++] = '#';
  for(int i = 0; i < n; ++i)  ma[l++] = s[i], ma[l++] = '#';
  ma[l] = 0;
  int mx = 0, id = 0;
  for(int i = 0; i < l; ++i){
    len[i] = mx > i ? min(len[2*id-i], mx-i) : 1;
    while(ma[i+len[i]] == ma[i-len[i]])  ++len[i];
    if(i + len[i] > mx){
      mx = i + len[i];  id = i;
    }
    lx[i-len[i]+1] = max(lx[i-len[i]+1], len[i]-1);
    rx[i+len[i]-1] = max(rx[i+len[i]-1], len[i]-1);
  }
  for(int i = 1; i < l; ++i)  lx[i] = max(lx[i], lx[i-1] - 1);
  for(int i = l-2; i >= 0; --i)  rx[i] = max(rx[i], rx[i+1] - 1);
}

int main(){
  scanf("%s", s);
  n = strlen(s);
  manacher(n);
  int ans = 0;
  for(int i = 0; i < (n<<1|1); ++i)  ans = max(ans, lx[i] + rx[i]);
  printf("%d
", ans);
  return 0;
}