POJ 3693 - Maximum repetition substring （后缀数组）

Maximum repetition substring

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7099 Accepted: 2118

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

ccabababcdaabbccaa#

Sample Output

Case 1: abababCase 2: aa

Source

2008 Asia Hefei Regional Contest Online by USTC

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题意：

给一个字符串，求一个重复次数最多的重复子串。

后缀数组经典应用，09年罗穗骞论文上的题目

他说了往后的匹配，但是没有说往前匹配的方法

#include <cstdio>#include <iostream>#include <vector>#include <algorithm>#include <cstring>#include <string>#include <map>#include <cmath>#include <queue>#include <set>using namespace std;//#define WIN#ifdef WINtypedef __int64 LL;#define iform "%I64d"#define oform "%I64d\n"#define oform1 "%I64d"#elsetypedef long long LL;#define iform "%lld"#define oform "%lld\n"#define oform1 "%lld"#endif#define S64I(a) scanf(iform, &(a))#define P64I(a) printf(oform, (a))#define P64I1(a) printf(oform1, (a))#define REP(i, n) for(int (i)=0; (i)<n; (i)++)#define REP1(i, n) for(int (i)=1; (i)<=(n); (i)++)#define FOR(i, s, t) for(int (i)=(s); (i)<=(t); (i)++)const int INF = 0x3f3f3f3f;const double eps = 1e-9;const double PI = (4.0*atan(1.0));const int maxn = 200000 + 20;char str[maxn];int s[maxn], sa[maxn], t[maxn], t2[maxn], c[maxn];void build_sa(int * s, int * sa, int n, int m) {    int i, *x = t, *y = t2;    for(i=0; i<m; i++) c[i] = 0;    for(i=0; i<n; i++) c[x[i] = s[i]]++;    for(i=1; i<m; i++) c[i] += c[i-1];    for(i=n-1; i>=0; i--) sa[--c[x[i]]] = i;    for(int k = 1; k <= n; k<<=1) {        int p = 0;        for(i=n-k; i<n; i++) y[p++] = i;        for(i=0; i<n; i++) if(sa[i] >= k) y[p++] = sa[i]-k;        for(i=0; i<m; i++) c[i] = 0;        for(i=0; i<n; i++) c[x[y[i]]]++;        for(i=0; i<m; i++) c[i] += c[i-1];        for(i=n-1; i>=0; i--) sa[--c[x[y[i]]]] = y[i];        swap(x, y);        p = 1; x[sa[0]] = 0;        for(i=1; i<n; i++)            x[sa[i]] = y[sa[i-1]]==y[sa[i]] && y[sa[i-1]+k]==y[sa[i]+k] ? p-1 : p++;        if(p >= n) break;        m = p;    }}int rank[maxn], height[maxn];void getHeight(int * s, int * sa, int n) {    int i, j, k = 0;    for(i=0; i<n; i++) rank[sa[i]] = i;    for(i=0; i<n; i++) {        if(k) k--;        int j = sa[rank[i]-1];        while(s[i+k] == s[j+k]) k++;        height[rank[i]] = k;    }}int RMQ_d[maxn][20];void RMQ_init(int * A, int n) {    for(int i=1; i<=n; i++) RMQ_d[i][0] = A[i];    for(int j=1; (1<<j) <= n; j++)        for(int i=1; i+j-1 <= n; i++)            RMQ_d[i][j] = min(RMQ_d[i][j-1], RMQ_d[i+(1<<(j-1))][j-1]);}int LCP(int L, int R) {    L = rank[L], R = rank[R];    if(L > R) swap(L, R);    L++;    int k = 0;    while((1<<(k+1)) <= R-L+1) k++;    return min(RMQ_d[L][k], RMQ_d[R-(1<<k)+1][k]);}int A[maxn];int main() {    int kase = 1;    while(scanf("%s", str) != EOF && str[0] != '#') {        int n = strlen(str);        for(int i=0; i<n; i++) s[i] = str[i] - 'a' + 1;        s[n] = 0;        build_sa(s, sa, n+1, 27);        getHeight(s, sa, n+1);        /*        for(int i=0; i<=n; i++) {            printf("%d (%02d) : ", height[i], sa[i]);            for(int j=sa[i]; j<n; j++) putchar('a'+s[j]-1);            putchar('\n');        }        */        // test n        RMQ_init(height, n+1);        int pos = 0, ans = 0, ansn = 0;        int cnt = 0;        for(int len=1; len<=n/2; len++) {            for(int i=0; i+len<n; i+=len) {                int tot = LCP(i, i+len);                int m = len - (tot%len);                m = i - m;                if(m >= 0 && m+len < n && tot%len) tot = max(tot, LCP(m, m+len));                tot = tot / len + 1;                if(tot > ans) cnt = 0;                if(tot >= ans) {                    ans = tot;                    A[cnt++] = len;                }            }        }        int len = 0;        int flag = true;        for(int i=1; i<=n && flag; i++) {            for(int j=0; j<cnt; j++) {                int tl = A[j];                if(LCP(sa[i], sa[i]+tl) >= (ans-1)*tl) {                    flag = false;                    pos = sa[i];                    len = tl * ans;                    break;                }            }        }        str[pos+len] = '\0';        printf("Case %d: %s\n", kase++, str+pos);    }    return 0;}