poj 3693

来源：互联网发布：农村假冒伪劣数据编辑：程序博客网时间：2024/05/29 15:52

题意：一个循环串是指能由其中某个子串循环出现若干次得到的字符串。求一个字符串中的循环次数最多的一个循环子串。

分析：假设要求的循环串是由子串s循环得到的，现在枚举s的长度L，则若循环次数是n，一定会覆盖s[0],s[L],s[2*L]......中相邻的n个点。取最左边的两个点s[i*L]和s[i*L+L]，求其最长公共前缀lcp(i*L,i*L+L)，若为K，则len=K/L+1为循环串的可能长度，之所以是”可能长度“，是因为枚举的s[i*L]完全可能不是循环串的开始字符，因此还要向前枚举字符，显然L-K%L是循环串开始字符与s[i*L]之间至少字符数量，此时再求lcp(i*L-(L-K%L),i*L+L-(L-K%L) )，若仍大于等于K，则使len++。

代码如下：

#include <cstdio>#include <stack>#include <set>#include <iostream>#include <string>#include <vector>#include <queue>#include <list>#include <functional>#include <cstring>#include <algorithm>#include <cctype>#include <string>#include <map>#include <iomanip>#include <cmath>#define LL long long#define ULL unsigned long long#define SZ(x) (int)x.size()#define MP(a, b) make_pair(a, b)#define MS(arr, num) memset(arr, num, sizeof(arr))#define PB push_back#define F first#define S second#define ROP freopen("input.txt", "r", stdin);#define MID(a, b) (a + ((b - a) >> 1))#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1#define lrt rt << 1#define rrt rt << 1|1#define root 1,n,1#define BitCount(x) __builtin_popcount(x)#define BitCountll(x) __builtin_popcountll(x)#define LeftPos(x) 32 - __builtin_clz(x) - 1#define LeftPosll(x) 64 - __builtin_clzll(x) - 1const double PI = acos(-1.0);const int INF = 1e7;using namespace std;const double eps = 1e-5;const int MAXN = 300 + 10;const int MOD = 1000007;const double M=1e-8;const int N=100100;typedef pair<int, int> pii;typedef pair<int, string> pis;int sa[N],rank[N],height[N];int cnt[N],wa[N],wb[N],wv[N],dp[N][33],a[N];bool cmp(int *y,int a,int b,int l){    return y[a]==y[b] && y[a+l]==y[b+l];}void da(char s[],int n,int m){    int i,j,p,*x=wb,*y=wa;    MS(cnt,0);    for (i=0;i<n;i++) cnt[x[i]=s[i]]++;    for (i=1;i<m;i++) cnt[i]+=cnt[i-1];    for (i=n-1;i>=0;i--) sa[--cnt[x[i]]]=i;    for (j=1,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<n;i++) wv[i]=x[y[i]];        MS(cnt,0);        for (i=0;i<n;i++) cnt[wv[i]]++;        for (i=1;i<m;i++) cnt[i]+=cnt[i-1];        for (i=n-1;i>=0;i--) sa[--cnt[wv[i]]]=y[i];        swap(x,y);        for (i=1,p=1,x[sa[0]]=0;i<n;i++) {            x[sa[i]]=cmp(y,sa[i],sa[i-1],j)?p-1:p++;        }    }}void getHeight(char s[],int n){    int i,j,k=0;    for (i=1;i<=n;i++) rank[sa[i]]=i;    for (i=0;i<n;height[rank[i++]]=k) { //cout<<"->"<<k<< " "<<i<<endl;        for (k?k--:0,j=sa[rank[i]-1];s[i+k]==s[j+k];k++) ;    }}void initRMQ(int n){    int i,j,t=log(n)/log(2);    for (i=1;i<=n;i++) dp[i][0]=height[i];    for (j=1;j<=t;j++) {        for (i=1;i+(1<<j)<=n;i++) dp[i][j]=min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);    }}int query(int x,int y){    if (x>y) swap(x,y);    x++;    int t=log(y-x+1)/log(2);    return min(dp[x][t],dp[y-(1<<t)+1][t]);}int lcp(int x,int y){    return query(rank[x],rank[y]);}int start,len;  // 开始位置，长度int slove(char s[], int n){    int i,j,e,k=n>>1,mx=-1,res,cnt2=0;    for (i=1;i<=k;i++) {        for (j=0;j<n-i;j+=i) if (s[j]==s[j+i]) {            int r=lcp(j,j+i);            res=r/i+1;            int t=j-(i-r%i);            if (t>=0 && r%i) {                if (lcp(t,t+i)>=r) res++;            }            if (res>mx) {                mx=res;                cnt2=0;                a[cnt2++]=i;            }            else if (res==mx) {                a[cnt2++]=i;            }        }    }           //  for (i=0;i<cnt2;cout<<a[i++]<<" - "); cout<<endl;    len=-1; // cout<<mx<<endl;    for (i=1;i<=n && len==-1;i++) {        for (j=0;j<cnt2;j++) {            int l=a[j];            if (lcp(sa[i],sa[i]+l)>=(mx-1)*l) {                len=l*mx;                start=sa[i];                break;            }        }    }    return mx;}int main(){    char s[N];    int n,i,j,K=1;    while(~scanf("%s",s))    {        if (s[0]=='#') break;        printf("Case %d: ",K++);        n=strlen(s);        if (n==1) {            puts(s);            continue;        }        s[n]=0;        da(s,n+1,128);        getHeight(s,n);        initRMQ(n);        slove(s,n);  //cout<<"->"<<start<<" "<<len<<endl;        for (i=start;i<start+len;i++) printf("%c",s[i]);        puts("");    }}/*cccabcabcccababababczzzzyykkllzzbbabbabbabbabbabbabbbbbbabbabbabccbabbabbab*/

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