POJ 3974 Palindrome 解题报告 回文串 manacher

POJ 3974 Palindrome 解题报告

manacher算法求最大回文串
解题思路：manacher算法。看上面链接就能理解了，注意链接里代码有错误地方，不要直接抄，自己改下就行了。

#include<iostream>
#include<math.h>
#include<iomanip>
#include<algorithm>
#include<iostream>
#include<math.h>
#include<iomanip>
#include<algorithm>
#include<queue>
#include<cstring>
#include<string>
#include<map>
#include<stack>
#include<stdio.h>
#include<cstdio>
#include<stdlib.h>
#include<fstream>
#include<iomanip>
#pragma warning(disable:4996)
#define INF 0x3f3f3f3f
#define ll long long
#define PI acos(-1.0)
const int N = 1000010;
const int maxn = 1e9;
using namespace std;
char str[2000005];
char a[1000005];
int len[2000005];
int main()
{
	int cnt = 0;
	while (scanf("%s",&a))
	{
		cnt++;
		if (strcmp(a, "END") == 0)
			break;
		str[0] = '@';//头和尾的字符都要初始化，且初始化为不一样的字符，防止越界
		int l = strlen(a);
		for (int i = 1; i <= 2 * l; i += 2)
		{
			str[i] = '#';
			str[i + 1] = a[i / 2];
		}
		str[2 * l + 1] = '#';
		str[2 * l + 2] = '!';//初始化尾部
		int rmax = 0, ans = 1, mid = 0;
		for (int i = 1; i <= l*2; i++)
		{
			if (rmax > i)
				len[i] = min(rmax - i, len[2 * mid - i]);//j+i=2*mid,len[i]在len[j]和rmax-i中取小
			else
				len[i] = 1;//如果不在已知回文串内，就要重新计算
			while (str[i - len[i]] == str[i + len[i]])
				len[i]++;
			if (i + len[i] > rmax)
			{
				rmax = i + len[i];//更新已知右边界
				mid = i;
			}
			ans = max(ans, len[i]-1);
		}
		printf("Case %d: %d\n", cnt,ans);
	}
}