poj3693之后缀数组
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 5946 | Accepted: 1799 |
Description
The repetition number of a string is defined as the maximum number R such that the string can be partitioned into R same consecutive substrings. For example, the repetition number of "ababab" is 3 and "ababa" is 1.
Given a string containing lowercase letters, you are to find a substring of it with maximum repetition number.
Input
The input consists of multiple test cases. Each test case contains exactly one line, which
gives a non-empty string consisting of lowercase letters. The length of the string will not be greater than 100,000.
The last test case is followed by a line containing a '#'.
Output
For each test case, print a line containing the test case number( beginning with 1) followed by the substring of maximum repetition number. If there are multiple substrings of maximum repetition number, print the lexicographically smallest one.
Sample Input
ccabababc daabbccaa #
Sample Output
Case 1: ababab Case 2: aa
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <iomanip>
#define INF 99999999
typedef long long LL;
using namespace std;
const int MAX=100000+10;
int *rank,r[MAX],sa[MAX],height[MAX],L[MAX];
int wa[MAX],wb[MAX],wm[MAX],f[MAX][32];
char s[MAX];
bool cmp(int *r,int a,int b,int l){
return r[a] == r[b] && r[a+l] == r[b+l];
}
void makesa(int *r,int *sa,int n,int m){
int *x=wa,*y=wb,*t;
for(int i=0;i<m;++i)wm[i]=0;
for(int i=0;i<n;++i)wm[x[i]=r[i]]++;
for(int i=1;i<m;++i)wm[i]+=wm[i-1];
for(int i=n-1;i>=0;--i)sa[--wm[x[i]]]=i;
for(int i=0,j=1,p=0;p<n;j=j*2,m=p){//j表示合并的子串长度
for(p=0,i=n-j;i<n;++i)y[p++]=i;//对第二关键字超出数组范围的子串排序
for(i=0;i<n;++i)if(sa[i]>=j)y[p++]=sa[i]-j;//对剩下子串根据第二关键字排序
for(i=0;i<m;++i)wm[i]=0;
for(i=0;i<n;++i)wm[x[y[i]]]++;
for(i=1;i<m;++i)wm[i]+=wm[i-1];
for(i=n-1;i>=0;--i)sa[--wm[x[y[i]]]]=y[i];
for(t=x,x=y,y=t,i=p=1,x[sa[0]]=0;i<n;++i){//求新的x,相当于rank,但是相同的子串要排名相同
x[sa[i]]=cmp(y,sa[i],sa[i-1],j)?p-1:p++;//判断子串suffix(sa[i])与suffix(sa[i-1])是否相同并且确定排名
}
}
rank=x;
}
/*在计算好height[rank[i]]后,对于height[rank[i+1]],如果sa[rank[i]-1]的首字母和sa[rank[i]]首字母相同
则rank[i+1]肯定在rank[sa[rank[i]-1]+1]后面,根据排名为a,b的子串的最长公共前缀为[a,b]中最小的
所以i+1和sa[rank[i+1]-1]的公共前缀>=height[rank[i]]-1即>=k-1;
如果sa[rank[i]-1]的首字母和sa[rank[i]]首字母不相同,则上一次的k就是0,所以无影响
*/
void calheight(int *r,int *sa,int n){
for(int i=0,j=0,k=0;i<n;height[rank[i++]]=k){
for(k?--k:0,j=sa[rank[i]-1];r[i+k] == r[j+k];++k);
}
}
void InitRMQ(int n){
for(int i=1;i<=n;++i)f[i][0]=height[i];//初始化从i开始区间长度为2^0的最值
int l=log(n*1.0)/log(2.0);//2*l<=n
for(int j=1;j<=l;++j){
for(int i=1;i+(1<<j)-1<=n;++i){//i+2^j-1<=n
f[i][j]=min(f[i][j-1],f[i+(1<<(j-1))][j-1]);//i+2^j-1 - (i+2^(j-1))+1=2^(j-1)
}
}
}
int LCP(int i,int j){//求rank[i]与rank[i]+1,ran[i]+1与rank[i]+2...的最长公共前缀中的最值,即height[rank[i]+1]~height[rank[j]]的最值
i=rank[i],j=rank[j];
if(i>j)swap(i,j);
++i;
int l=log(j-i+1.0)/log(2.0);//2^l<=j-i+1
return min(f[i][l],f[j-(1<<l)+1][l]);
}
int main(){
int Case=0;
while(scanf("%s",s),s[0] != '#'){
int n=0;
for(n=0;s[n] != '\0';++n)r[n]=s[n];
r[n]=0;
makesa(r,sa,n+1,256);
calheight(r,sa,n);
InitRMQ(n);
int sum=0,size=0,x=sa[1],y=sa[1]+1;
for(int j=1;j<=n;++j){//对于长度为j的循环节,sum记录循环次数
for(int i=0;i+j<n;i+=j){
if(s[i] == s[i+j]){
int len=LCP(i,i+j);//向后匹配
int num=len/j;
int k=i-(j-len%j);
if(k>=0 && len%j && LCP(k,k+j)>=len)++num;//向前匹配
if(num == sum)L[++size]=j;//L记录得到最多循环次数的可能的子串长度
else if(num > sum)sum=num,L[size=0]=j;
}
}
}
for(int i=1;i<=n && sum;++i){//求哪个子串可以循环sum次
for(int j=0;j<=size;++j){
if(sa[i]+L[j]>=n)continue;
int len=LCP(sa[i],sa[i]+L[j]);
if(len/L[j] == sum){x=sa[i],y=sa[i]+(sum+1)*L[j],sum=0;break;}
}
}
printf("Case %d: ",++Case);
for(int i=x;i<y;++i)printf("%c",s[i]);
printf("\n");
}
return 0;
}