HDU 3294 马拉车算法

该算法，对于任意一个在最后 串中的pos位置，都对应原串的(pos-malache[pos]+2)/2-1位置，且长度为为malache[pos]-1

这样就很容易直接求出所有回文位置了。

模板依旧用上一篇文章的模板

/*
#include <bits/stdc++.h>
#include <ext/pb_ds/priority_queue.hpp>
#include <tr1/unordered_map>
#include <ctime>
using std::tr1::unordered_map;
*/

#include <iostream>
#include <cstring>
#include <queue>
#include <algorithm>
#include <cstdio>
#include <map>
using namespace std;

/*
using std::sort;
using std::bitset;
using std::max;
using std::cout;
using std::stack;
using std::cin;
using std::endl;
using std::swap;
using std::pair;
using std::vector;
using std::set;
using std::map;
using std::multiset;
using std::queue;
using std::greater;
using std::string;
using std::priority_queue;
using std::max_element;
using std::min_element;

using __gnu_pbds::pairing_heap_tag;
__gnu_pbds::priority_queue<int, greater<int>, pairing_heap_tag> heap;
#define Hash unordered_map
*/
#define pr(x) cout<<#x<<" = "<<x<<" "
#define prln(x) cout<<#x<<" = "<<x<<endl

#define lson o*2, L, M
#define rson o*2+1, M + 1,R

//const int maxn = 210000;
const int maxn = 400000+100;

//对于text[]，返回一个out[],out[i]，表示从i开始，可以向右多远是合法的

#include <typeinfo>
template<typename T>
void manacher(T text[], int len, int malache[], T str[])
{
T inf, space;
if (typeid(T).name() == typeid(char).name())
{
inf= -128;
space = 2 - 127;
}
if (typeid(T).name() == typeid(int).name())
{
inf= - 0x3f3f3f3f;
space = 2 -0x3f3f3f3f;
}
int l = 0;
str[l++] = inf;
str[l++] = space;
for (int i = 0; i < len; ++ i)
{
str[l++] = text[i];
str[l++] = space;
}
str[l] = inf + 1;
int mx=0, id=0;//mx:最远匹配距离,id:最远距离对应的坐标的下标
for (int i = 0; i < l; ++ i)
{
malache[i] = mx > i ? min(malache[2 * id - i], mx - i) : 1;
while (str[i + malache[i]] == str[i - malache[i]]) malache[i] ++ ;
if (i + malache[i] > mx)
{
mx = i + malache[i];
id = i;
}
}
}

char flag[10],input[400000 + 200];
int malache[400000 + 200];
char str[400000 + 200];

int main()
{
while (scanf(" %s %s ", flag, input) == 2)
{
char ch = flag[0];
int len = strlen(input);
for (int i = 0; i != len; ++ i)
{
input[i] = (input[i]-ch+26)%26 + 'a';
}
manacher(input, len, malache, str);
int pos=0, ans=0;
for (int i = 0; i < 2 * len + 2; i ++)
{
if (malache[i]>ans)
{
ans = malache[i];
pos = i;
}
}
ans--;
if (ans==1)
printf("No solution!\n");
else
{
if (pos%2==0)
{
//如果是偶数，则为pos/2-1为中点
//pos-malache[pos]+1为起点
//起点为 (pos-malache[pos]+1)/2-1为起点，长为malache[pos]-1
printf("%d %d\n", (pos-malache[pos]+2)/2-1, (pos-malache[pos]+2)/2-1+ans-1);
}
else
{
//如果是奇数，则说明为aa或者baab这样的回文串
//(pos-malache[pos])/2为起点，长度为malache[pos]-1
printf("%d %d\n", (pos-malache[pos]+2)/2-1, (pos-malache[pos]+2)/2-1+ans-1);
}
for (int i = (pos-malache[pos])/2; i <= (pos-malache[pos])/2 + ans - 1; ++ i)
printf("%c", input[i]);
printf("\n");
}
}
}