LightOJ 1258 Making Huge Palindromes (Manacher)

题意：给定上一个串，让你在后面添加一些字符，使得这个串成为一个回文串。

析：先用manacher算法进行处理如果发现有字符匹配超过最长的了，结束匹配，答案就是该字符前面那个长度加上该串原来的长度。

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>
#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 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 LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 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 t[maxn], s[maxn*2];
int p[maxn*2];

int main(){
  int T;  cin >> T;
  for(int kase = 1; kase <= T; ++kase){
    scanf("%s", t);
    memset(p, 0, sizeof p);
    s[0] = '$';  s[1] = '#';
    int cnt = 1;
    n = 0;
    for(int i = 0; t[i]; ++i, ++n){
      s[++cnt] = t[i];
      s[++cnt] = '#';
    }
    s[++cnt] = 0;
    int mx = 0, id = 0;
    int ans = 0;
    for(int i = 1; s[i]; ++i){
      p[i] = mx > i ? min(p[2*id-i], mx-i) : 1;
      while(s[p[i]+i] == s[i-p[i]])  ++p[i];
      if(p[i] + i > mx){
        mx = p[i] + i;
        id = i;
      }
      if(mx + 1 >= cnt){
        ans = (i - p[i]) / 2 + n;
        break;
      }
    }
    printf("Case %d: %d\n", kase, ans);
  }
  return 0;
}