LOJ 1258 - Making Huge Palindromes（KMP | Manacher）

本文链接： https://blog.csdn.net/qq_43408238/article/details/102722930

题意：给你一个字符串，然后让你添加最少的字符使得这个字符串变为一个回文串，只能在尾部添加。

思路：因为只能在尾部添加，所以我们只需要求一个最长的回文后缀，然后把这个后缀之前的字符添加到尾部就可以实现添加最少的字符使得这个串变为一个回文串。最后的长度 = 2 * len - ans (len为字符串的长度，ans为最长回文后缀的长度，考虑容斥原理，现在尾部添加这个字符串的翻转字符串，这个字符串一定回文，然后减去重复出现的回文后缀。）

AC Code：

KMP

#include<iostream>
#include<cstring>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<cmath>
#include<cstdio>
#include<iomanip>
#include<sstream>
#include<algorithm>

using namespace std;
#define read(x) scanf("%d",&x)
#define Read(x,y) scanf("%d%d",&x,&y)
#define sRead(x,y,z)  scanf("%d%d%d",&x,&y,&z)
#define gc(x)  scanf(" %c",&x);
#define mmt(x,y)  memset(x,y,sizeof x)
#define write(x) printf("%d\n",x)
#define INF 0x3f3f3f3f
#define ll long long
#define mod  998244353
#define pdd pair<double,double>
const int N = 1e6+5;
int Next[N];
char s[N];
char t[N];
int f[N];
void kmp_pre(int m){
    int i = 0,j = Next[0] = -1;
    while(i < m){
        while(j!= -1&& s[i]!= s[j]) j = Next[j];
        Next[++i] = ++j;

    }
}
void kmp(int n,int m){
    kmp_pre(m);
    int i = 0,j = 0;
    while(i < n){
        while(j != -1&&t[i] != s[j]) j = Next[j];
        f[++i] = ++j;
        if(f[i] >= m) j = Next[j];
    }
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    #endif // ONLINE_JUDGE

    int T;
    read(T);
    for(int p = 1; p<= T;++p){
        mmt(Next,0);
        mmt(f,0);
        scanf("%s",t);
        strcpy(s,t);
        int n = strlen(s);
        reverse(s , s + n);//翻转过来当做模式串，原串当做文本串
        kmp(n,n);//求出最长的后缀
        int ans = f[n];
        printf("Case %d: ",p);
        cout<<2*n - ans<<endl;
    }
}

Manacher

AC Code:

#include<iostream>
#include<cstring>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<cmath>
#include<cstdio>
#include<iomanip>
#include<sstream>
#include<algorithm>

using namespace std;
#define read(x) scanf("%d",&x)
#define Read(x,y) scanf("%d%d",&x,&y)
#define sRead(x,y,z)  scanf("%d%d%d",&x,&y,&z)
#define gc(x)  scanf(" %c",&x);
#define mmt(x,y)  memset(x,y,sizeof x)
#define write(x) printf("%d\n",x)
#define INF 0x3f3f3f3f
#define ll long long
#define mod  998244353
#define pdd pair<double,double>
const int N = 1000;
const int M=  2e6+5;
char Ma[M];
int Mp[M];
void Manacher(char s[],int len){
    int l = 0;
    Ma[l++] = '$';
    Ma[l++] = '#';
    for(int i  =0;i < len;++i){
        Ma[l++] = s[i];
        Ma[l++] = '#';
    }
    Ma[l ] = 0;
    int mx = 0,id = 0;
    for(int i = 0;i < l;++i){
        Mp[i] = mx > i ? min(Mp[2*id - i],mx - i):1;
        while(Ma[i + Mp[i]] == Ma[i - Mp[i]]) Mp[i]++;
        if(i + Mp[i] > mx){
            mx = i + Mp[i];
            id = i;
        }
    }
}
char s[M];
int main()
{
    int T;
    read(T);
    for(int p = 1;p <= T;++p){
        mmt(Ma,0);
        mmt(Mp,0);
        scanf("%s",s);
        int len = strlen(s);
        Manacher(s,len);
        int ans = 0;
        for(int i = 0;i < 2*len +2; ++i){
            int M = Mp[i] - 1;
            if(i + M == 2*len + 1) ans = max(ans,M);
        }
        printf("Case %d: %d\n",p,2*len - ans);
    }

}

LOJ 1258 - Making Huge Palindromes（KMP | Manacher）

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