HDU 3068 &&HDU 3294 +最长回文串*3—— manacher/扩展KMP/DP

HDU 3068 http://acm.hdu.edu.cn/showproblem.php?pid=3068

HDU 3294http://acm.hdu.edu.cn/showproblem.php?pid=3294

扩展KMP：https://segmentfault.com/a/1190000008663857

感觉DP相对实现起来更简洁一些，很容易想到，可以用bool dp[i][j]:表示从i到j是否为回文串，然后最长长度用一个变量记录就好

如果dp[i][j]为回文串，那么dp[i+1][j-1]必为回文串且s[i]==s[j]，所以可以得到递推关系

if(dp[i+1]][j-1]&&s[i]==s[j])  dp[i][j]=1;  maxlen=max(maxlen,i-j+1);

但是上述判断条件并不对……上面的条件漏掉了一个单个字符必为回文串的情况，第二个判断条件再加一个||j-i<2即可，或者最开始初始化dp数组的时候把i==j的部分赋值为true；

模板模板↓↓↓

bool  dp[n][n];
string s;
int n;
int DP() {
    memset(dp, 0, sizeof(dp));
    int maxlen = 0;
    for (int j = 0; j < s.length(); j++)
        for (int i = j; i >= 0; i--)
            if(s[i]==s[j]&&(j-i<2|| dp[i + 1][j - 1]))
            {
                dp[i][j] = 1;
                maxlen = max(maxlen, j - i + 1);
            }
    return maxlen;
}

但是……dp数组这样开就太大了……相对短一些的字符串还能用，长的就不行了……

manacher

manacher：https://segmentfault.com/a/1190000003914228#articleHeader8

HDU 3086

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int MAX = 110000 + 10;
char s[MAX*2];//数组要开两倍的
int rl[MAX*2];
int n,pos,maxlen,maxright;
int manacher()
{
    for (int i = n; i >= 0; i--)
    {
        s[2 * i + 2] = s[i];
        s[2 * i + 1] = ‘#‘;
    }
    s[0] = ‘*‘;
    s[n * 2 + 2] = ‘\0‘;
    //大佬的小技巧，防止数组越界
    for (int i = 2; i < 2 * n + 1; i++)
    {
        if (maxright > i)  //i在pos左边
            rl[i] = min(rl[2 * pos-i], maxright- i);//i在pos右边
        else rl[i] = 1;
        while (s[i - rl[i]] == s[i + rl [i]]) rl[i]++; //向两边扩展
        if (maxright < rl[i]+i - 1)  //更新maxright
        {
            maxright = rl[i] + i - 1; pos = i;
        }
        if (maxlen < rl[i]) maxlen = rl[i];
    }
    return maxlen-1;
}
int main()
{
    while (scanf("%s", s)!=EOF)
    {
        n = strlen(s); maxright = 0;
        pos = 0; maxlen = 0;
        printf("%d\n",manacher());
    }
    return 0;
}

HDU3298

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int MAX = 200000 + 10;
int rl[MAX * 2];
char s[MAX * 2],t[MAX*2], a[5];
int len, maxlen, pos, st,maxright,ed;
void manacher()
{
    s[0] = ‘*‘; s[2 * len + 2] = ‘\0‘;
    for (int i = 0; i < len; i++)
    {
        s[i * 2 + 1] = ‘#‘;
        s[i * 2 + 2] = t[i];
    }
    for (int i = 2; i < 2 * len + 2; i++)
    {
        if (i < maxright) rl[i] = min(maxright - i, rl[2 * pos - i]);
        else rl[i] = 1;
        while (s[i + rl[i]] == s[i - rl[i]]) rl[i]++;
        if (i + rl[i]-1 > maxright)
        {
            maxright = i + rl[i]-1;
            pos = i;
        }
        if (maxlen < rl[i])
        {
            maxlen = rl[i];
            st =(i-rl[i])/2;
            ed = (i + rl[i]) / 2-2;
        }
    }
    maxlen--;
}

int main()
{
    while (scanf("%s%s",a,t)!= EOF)
    {
        //getchar();
        len = strlen(t);
        for (int i = 0; i < len; i++)
        {
            int b = t[i] - a[0];
            if (b < 0) b += 26;
            t[i] = b + ‘a‘;
        }
        maxlen = 0;
        maxright = 0; pos = 0;
        st = 0; ed = 0;
        manacher();
        if (maxlen==1)printf("No solution!");
        else
        {
            printf("%d %d\n", st, ed);
            for (int i = st; i <= ed; i++)
                printf("%c",t[i] );
        }
        puts("");
    }
  return 0;
}

扩展KMP

emmmmmmm……还不太会……先放着，会了再更