2015
04-14

The Number of Palindromes

Now, you are given a string S. We want to know how many distinct substring of S which is palindrome.

The first line of the input contains a single integer T(T<=20), which indicates number of test cases.
Each test case consists of a string S, whose length is less than 100000 and only contains lowercase letters.

The first line of the input contains a single integer T(T<=20), which indicates number of test cases.
Each test case consists of a string S, whose length is less than 100000 and only contains lowercase letters.

3
aaaa
abab
abcd

Case #1: 4
Case #2: 4
Case #3: 4

UESTC的神题啊！

pre2同理更新。

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
#define MAXN 200100
int dp[MAXN][20];
char r[MAXN];
int sa[MAXN];
int wa[MAXN],wb[MAXN],wv[MAXN],ws[MAXN];
int height[MAXN],rk[MAXN];
int mm[MAXN];
int cas;
inline bool cmp(int *r,int a,int b,int len){
return r[a]==r[b]&&r[a+len]==r[b+len];
}
void SA(int n,int m){
int i,j,p,*x=wa,*y=wb,*t;
for(i=0;i<m;i++)
ws[i]=0;
for(i=0;i<n;i++)
ws[x[i]=r[i]]++;
for(i=1;i<m;i++)
ws[i]+=ws[i-1];
for(i=n-1;i>=0;i--)
sa[--ws[x[i]]]=i;
for(j=p=1;p<n;j<<=1,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<m;i++)
ws[i]=0;
for(i=0;i<n;i++)
ws[wv[i]=x[y[i]]]++;
for(i=1;i<m;i++)
ws[i]+=ws[i-1];
for(i=n-1;i>=0;i--)
sa[--ws[wv[i]]]=y[i];
for(t=x,x=y,y=t,x[sa[0]]=0,p=i=1;i<n;i++)
x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
}
}
void Height(int n){
int i,j,k=0;
for(i=0;i<=n;i++)    //这里sa[0]为‘\0’开始的子串
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 init(){
int i,j,len;
memset(height,0,sizeof(height));
len=strlen(r);
r[len]='\$';
for(j=0;j<len;j++)
r[j+len+1]=r[len-1-j];
r[j+len+1]='\0';
return len*2+1;
}
void st(int n){
int i,j,p,q;
for(i=1;i<=n;i++)
dp[i][0]=i;
for(j=1;j<=mm[n];j++)
for(i=1;i+(1<<j)-1<=n;i++){
p=height[dp[i][j-1]];
q=height[dp[i+(1<<(j-1))][j-1]];
if(p>q)
dp[i][j]=dp[i+(1<<(j-1))][j-1];
else
dp[i][j]=dp[i][j-1];
}

}
int RMQ_MIN(int i,int j){
int tem;
if(i>j){
tem=i;
i=j;
j=tem;
}
i++;           //交换后小的要加一
int k=mm[(j-i+1)];
return min(height[dp[i][k]],height[dp[j-(1<<k)+1][k]]);
}
void solve(int n){
int i,j,ans=0,pre1=0,pre2=0;
st(n);
for(i=2;i<=n;i++){
pre1=min(pre1,height[i]);
j=RMQ_MIN(i,rk[n-1-sa[i]]);
if(j>pre1){
ans+=j-pre1;
pre1=j;
}
pre2=min(pre2,height[i]);
j=RMQ_MIN(i,rk[n-sa[i]]);
if(j>pre2){
ans+=j-pre2;
pre2=j;
}
}
printf("Case #%d: %d\n",cas,ans);
}
int main(){
int i,j,n,t,T;
mm[0]=-1;
for(i=1;i<MAXN;i++)
mm[i]=((i&(i-1))==0)?mm[i-1]+1:mm[i-1];           //这里预处理一下，时间跑快了1s

scanf("%d",&T);
cas=0;
for(t=1;t<=T;t++){
scanf("%s",r);
n=init();
SA(n+1,130);
Height(n);
cas++;
solve(n);
}
}