Codeforces Global Round 7 cf 1326D 馬拉車

題意是叫你找最長的迴文串，然後這個迴文串可以由原本的串的前驅+後繼拼起來。

由於前驅和後繼是必須選的對不對（可能有一個爲空）。
那麼我們第一步就是吧原本的串倒置，把正的和反的去找出最長的相同的，這樣就滿足了前驅+後繼的限制。下一步是什麼？

假設原本的長度是[1,len]，我們上面找到的最長相同的長度是3，那麼就是接下來剩下的串就是[1 + 3 - 1 , len - 3],剩下的這一段我們只需要找到兩個端點爲開頭的最長迴文串就好了。如圖所示。

爲什麼是兩個端點？如果選取了左端點則說明我們與前驅的串接上去了。如果選擇了右端點，這說明我們與後繼的串接上去了。這樣就滿足了題意。要求這個的長度我用了馬拉車。複雜度是O(N)，所以這道題也就解決了。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<time.h>
#include<string>
#include<cmath>
#include<stack>
#include<map>
#include<set>
#define int long long
#define double long double
using namespace std;
#define PI  3.1415926535898
#define eqs 1e-10
const long long max_ = 1e6 + 7;
const int mod = 998244353;
const int inf = 1e9 + 7;
const long long INF = 1e18;
int read() {
	int s = 0, f = 1;
	char ch = getchar();
	while (ch<'0' || ch>'9') {
		if (ch == '-')
			f = -1;
		ch = getchar();
	}
	while (ch >= '0'&&ch <= '9') {
		s = s * 10 + ch - '0';
		ch = getchar();
	}
	return s * f;
}
inline int min(int a, int b) {
	return a < b ? a : b;
}
inline int max(int a, int b) {
	return a > b ? a : b;
}
int nn = 1, pr[max_ * 2], weizhi, extralen = -1;
char node[2 * max_],yuan[max_];
void dr(int L,int R) {
	nn = 1; extralen = -1;
	node[0] = '~';
	node[1] = '|';
	for(int i = L;i <= R;i++) {
		node[++nn] = yuan[i];
		node[++nn] = '|';
	}
	int mid = 0, maxr = 0;
	for (int i = 1; i <= nn; i++) {
		if (i <= maxr && i >= mid) {
			pr[i] = min(pr[mid * 2 - i], maxr - i);
		}
		else pr[i] = 1;
		while (node[i + pr[i]] == node[i - pr[i]]) {
			pr[i]++;
		}
		if (pr[i] + i > maxr) {
			maxr = pr[i] + i;
			mid = i;
		}
		if (pr[i] - 1 >= extralen &&
			(i - pr[i] == 0 || i + pr[i] == nn 
				|| i - pr[i] == 1 || i + pr[i] == nn + 1) ) {
			extralen = pr[i] - 1; 
			weizhi = i;
		}
	}
}
signed main() {
	std::ios::sync_with_stdio(false);
	int T;
	cin >> T;
	while (T--){
		cin >> yuan + 1;
		int shouweilen = 0,len = strlen(yuan + 1);
		for (int i = 1, j = len; i <= len; i++, j--) {
			if (yuan[i] == yuan[j]) {
				shouweilen = i;
			}
			else break;
		}
		for (int i = 1; i <= shouweilen; i++) {
			cout << yuan[i];
		}
		if (shouweilen == len) {
			cout << "\n"; continue;
		}
		dr(shouweilen + 1, len - shouweilen);
		/*for (int i = 1; i <= nn; i++) {
			cout << "i = "<< i<<"  :"<<node[i] << " " << pr[i]<<endl;
		}*/
		for (int i = weizhi - pr[weizhi] + 1; i < weizhi + pr[weizhi]; i++) {
			if (node[i] == '|')continue;
			else cout << node[i];
		}
		for (int i = len - shouweilen + 1; i <= len ; i++) {
			cout << yuan[i];
		}
		cout << "\n";
		for (int i = 0; i <= nn + 2; i++)
			pr[i] = 0, node[i] = '\0';
	}
	return 0;
}