hdu3068&&hdu3294，回文串，Manacher算法

关于Manacher算法讲解在这
在o（n）时间内算出以每个点为中心的最大回文串长度

hdu3068
模板题

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N=233333;//20W

int Manacher(string st){
    int len=st.size();
    int *p=new int[len+1];
    memset(p,0,sizeof(p));
    int mx=0,id=0;
    for (int i=1;i<=len;i++){
        if (mx>i)p[i]=min(p[2*id-i],mx-i);
        else p[i]=1;
        while (st[i+p[i]]==st[i-p[i]])p[i]++;
        if (i+p[i]>mx){mx=i+p[i];id=i;}
    }
    int ma=0;
    for(int i=1;i<len;i++)ma=max(ma,p[i]);
    delete(p);
    return ma-1;
}

int main(){
    //freopen("fuck.in","r",stdin);
    char st[N];
    while (~scanf("%s",st)){
        string st0="$#";
        for (int i=0;st[i]!='\0';i++){
            st0+=st[i]; st0+="#";
        }
        printf("%d\n",Manacher(st0));
    }
    return 0;
} 

hdu3294
最后输出的时候稍微处理一下就好

#include<map>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N=200400;
string ans;
int _start,_end;

int Manacher(string st){
    int len=st.size();
    int *p=new int[len+1];
    memset(p,0,sizeof(p));
    int mx=0,id=0;
    for (int i=1;i<=len;i++){
        if (mx>i)p[i]=min(p[id*2-i],mx-i);
        else p[i]=1;
        while (st[i+p[i]]==st[i-p[i]])p[i]++;
        if (i+p[i]>mx)mx=i+p[i],id=i;
    }
    int ma=0,ii;
    for (int i=1;i<len;i++)
        if (p[i]>ma)ma=p[i],ii=i;
    ma--;
    _start=((ii-ma+1)>>1)-1;
    _end  =((ii+ma-1)>>1)-1;
    ans="";
    for (int i=ii-ma;i<=ii+ma;i++)
        if (st[i]!='#')ans+=st[i];
    delete(p);
    return ma;
}

void writeln(char ch,string st){
    char tag='a'-1;
    map<char,char>t;
    for (char i=ch;i<='z';i++)t[i]=++tag;
    for (char i='a';i< ch;i++)t[i]=++tag;
    int len=st.size();
    printf("%d %d\n",_start,_end);
    for (int i=0;i<len;i++)printf("%c",t[st[i]]);
    puts("");
}

int main(){
    char ch,st[N];
    while (~scanf("%c %s\n",&ch,st)){
        string st0="$#";
        for (int i=0;st[i]!='\0';i++)
            {st0+=st[i];st0+="#";}
        if (Manacher(st0)==1)puts("No solution!");
        else writeln(ch,ans);
    }
    return 0;
} 

明天高数月考，，，，求rp++