Girls' research HDU - 3294(马拉车水题)

Posted 2020-12-30 wtsruvf

题意：

　　求最长回文串 长度要大于等于2  且输出起点和终点  输出回文串字符 这个字符还是要以给出的字符为起点a 输出

解析：

　　分析一下s_new串就好了

　　

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 200010, INF = 0x7fffffff;
char s[maxn], s_new[maxn<<1];
int p[maxn<<1];
int tmp;
int init()
{
    int len = strlen(s);
    s_new[0] = ‘$‘;
    s_new[1] = ‘#‘;
    int j = 2;
    for(int i=0; i<len; i++)
        s_new[j++] = s[i], s_new[j++] = ‘#‘;
    s_new[j++] = ‘‘;
    return j;
}

int manacher()
{
    int len = init();
    int max_len = -1;
    int id, mx = 0;
    for(int i=1; i<len; i++)
    {
        if(i < mx)
            p[i] = min(p[2*id - i], mx - i);
        else
            p[i] = 1;
        while(s_new[i-p[i]] == s_new[i+p[i]])
            p[i]++;
        if(mx < i+p[i])
        {
            mx = i+p[i];
            id = i;
        }
        if(max_len < p[i] - 1)
            max_len = p[i] - 1, tmp = i;
    }
    return max_len;
}


char c;
int main()
{
    while(~scanf("%c%s", &c, s))
    {
        getchar();
        int max_len = manacher();
        if(max_len < 2)
            printf("No solution!
");
        else
        {
            int a = (tmp-p[tmp])/2, b = (tmp-p[tmp])/2+p[tmp]-2;
            printf("%d %d
", a, b);
            for(int i=a; i<=b; i++)
            {
                if(s[i] >= c)
                    printf("%c", s[i] - c + ‘a‘);
                else
                    printf("%c", ‘z‘ - c + s[i] + 1);
            }
            printf("
");


        }
    }


    return 0;
}