UVA 11475 Extend to Palindrome (kmp || manacher || 后缀数组)

题目链接：点击打开链接

题意：给你一个串，让你在串后面添加尽可能少的字符使得这个串变成回文串。

思路：这题可以kmp,manacher,后缀数组三种方法都可以做，kmp和manacher效率较高，时间复杂度是O(n)，后缀数组时间复杂度是O(nlogn).思路是求出元串的后缀和反串的前缀匹配的最大长度。用后缀数组的时候求出l=lcp(i,len+1)，判断l+i是不是等于len,如果等于那么就是结果。

kmp:

#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#include<vector>#include<map>#include<set>#include<string>#include<bitset>#include<algorithm>using namespace std;#define lson th<<1#define rson th<<1|1typedef long long ll;typedef long double ldb;#define inf 99999999#define pi acos(-1.0)#define maxn 1000600char s1[maxn],s2[maxn];int nextt[maxn],len;int pd(char *s1,char *s2){    int i,j,len1,len2;    len1=len2=len;    i=0;j=-1;    memset(nextt,-1,sizeof(nextt));    while(i<len2){        if(j==-1 || s2[i]==s2[j]){            i++;            j++;            nextt[i]=j;        }        else j=nextt[j];    }    i=0;j=0;    int ans=1;    while(i<len1 && j<len2){        if(j==-1 || s1[i]==s2[j]){            //ans=max(ans,j+1);  注意这句不能加，因为要求的是当i=len1时候，串二最大匹配了多少长度            i++;            j++;        }        else j=nextt[j];   //这里要解释一下，这里的意思是s[i]和s[j]不匹配了，那么看前j个字符前缀和后缀匹配的程度，注意，这里前j个字符是s[0]...s[j-1]    }    ans=j;    return ans;}int main(){    int n,m,i,j,ans;    while(scanf("%s",s1)!=EOF)    {        strcpy(s2,s1);        len=strlen(s1);        reverse(s2,s2+len);        ans=pd(s1,s2);        printf("%s",s1);        for(i=len-1-ans;i>=0;i--){            printf("%c",s1[i]);        }        printf("\n");    }    return 0;}

manacher:

#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#include<vector>#include<map>#include<set>#include<string>#include<bitset>#include<algorithm>using namespace std;#define lson th<<1#define rson th<<1|1typedef long long ll;typedef long double ldb;#define inf 99999999#define pi acos(-1.0)#define maxn 100060char s[maxn],ss[maxn*2];  //要开两倍大小int p[maxn*2];int main(){    int n,m,i,j,mx,idx,maxx,len,len1;    while(scanf("%s",s)!=EOF)    {        len=strlen(s);        len1=len;        ss[0]='$';        ss[1]='#';        for(i=0;i<len;i++){            ss[i*2+2]=s[i];            ss[i*2+3]='#';        }        mx=0;maxx=0;        len=2*len+2;        int ans=inf;        for(i=1;i<len;i++){            if(mx>i){       //这里的mx为满足条件的后一个，这样如果i<=idx+p[i]-1就能比较了                p[i]=min(p[idx*2-i],mx-i);            }            else p[i]=1;            while(ss[i+p[i]]==ss[i-p[i]]){                p[i]++;            }            if(maxx<p[i])maxx=p[i];            if(mx<i+p[i]){  //这里更新的mx也是范围内的后一个                mx=i+p[i];                idx=i;            }            if(i+p[i]-1==len-1 ){                //printf("--->%d\n",i);                ans=min(ans,(i-(p[i]-1)) /2 );            }        }        printf("%s",s);        for(i=ans-1;i>=0;i--){            printf("%c",s[i]);        }        printf("\n");        //printf("%d\n",ans);    }    return 0;}

后缀数组：

#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#include<vector>#include<map>#include<set>#include<string>#include<bitset>#include<algorithm>using namespace std;#define lson th<<1#define rson th<<1|1typedef long long ll;typedef long double ldb;#define inf 99999999#define pi acos(-1.0)#define M 100050#define maxn 200050char s1[M],s2[M];int a[maxn];int idx(char c){    if(c>='a' && c<='z')return c-'a'+1;    if(c>='A' && c<='Z')return c-'A'+1+26;}int sa[maxn];int wa[maxn],wb[maxn],wv[maxn],we[maxn];int rk[maxn],height[maxn];int cmp(int *r,int a,int b,int l){    return r[a]==r[b]&&r[a+l]==r[b+l];}void build_sa(int *r,int n,int m){    int i,j,p,*x=wa,*y=wb,*t;    for(i=0;i<m;i++)we[i]=0;    for(i=0;i<n;i++)we[x[i]=r[i]]++;    for(i=1;i<m;i++)we[i]+=we[i-1];    for(i=n-1;i>=0;i--)sa[--we[x[i]]]=i;    for(j=1,p=1;p<n;j*=2,m=p){        for(p=0,i=n-j;i<n;i++)y[p++]=i;        for(i=0;i<n;i++)if(sa[i]>=j)y[p++]=sa[i]-j;        for(i=0;i<n;i++)wv[i]=x[y[i]];        for(i=0;i<m;i++)we[i]=0;        for(i=0;i<n;i++)we[wv[i]]++;        for(i=1;i<m;i++)we[i]+=we[i-1];        for(i=n-1;i>=0;i--)sa[--we[wv[i]]]=y[i];        for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1;i<n;i++)        x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;    }}void calheight(int *r,int n){    int i,j,k=0;    for(i=1;i<=n;i++)rk[sa[i]]=i;    for(i=0;i<n;height[rk[i++] ]=k){        for(k?k--:0,j=sa[rk[i]-1];r[i+k]==r[j+k];k++);    }}int minx[200100][30];void init_rmq(int n){    int i,j;    for(i=1;i<=n;i++)minx[i][0]=height[i];    for(j=1;j<=20;j++){        for(i=1;i<=n;i++){            if(i+(1<<j)-1<=n)            {                minx[i][j]=min(minx[i][j-1],minx[i+(1<<(j-1))][j-1]);            }        }    }}int lcp(int l,int r){    int k,i;    l=rk[l];r=rk[r];    if(l>r)swap(l,r);    l++;    k=(log((r-l+1)*1.0)/log(2.0));    return min(minx[l][k],minx[r-(1<<k)+1][k]);}int main(){    int n,m,i,j,len,l;    while(scanf("%s",s1)!=EOF)    {        strcpy(s2,s1);        len=strlen(s1);        reverse(s2,s2+len);        n=0;        for(i=0;i<len;i++){            a[n++]=idx(s1[i]);        }        a[n++]=55;        for(i=0;i<len;i++){            a[n++]=idx(s2[i]);        }        a[n]=0;        build_sa(a,n+1,60);        calheight(a,n);        init_rmq(n);        for(i=0;i<len;i++){            l=lcp(i,len+1);            if(i+l==len)break;        }        printf("%s",s1);        for(i=len-1-l;i>=0;i--){            printf("%c",s1[i]);        }        printf("\n");    }    return 0;}