poj 3693 Maximum repetition substring(后缀数组)

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题目链接：poj 3693 Maximum repetition substring

题目大意：求一个字符串中循环子串次数最多的子串。

解题思路：对字符串构建后缀数组，然后枚举循环长度，分区间确定。对于一个长度l，每次求出i和i+l的LCP，那么以i为起点，循环子串长度为l的子串的循环次数为LCP/l+1，然后再考虑一下从i-l+1~i之间有没有存在增长的可能性。

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;const int maxn = 100005;struct Suffix_Arr {    int n, s[maxn];    int SA[maxn], rank[maxn], height[maxn];    int tmp_one[maxn], tmp_two[maxn], c[305];    int d[maxn][20];    void init(char* str);    void build(int m);    void get_height();    void rmq_init();    int rmq_query(int x, int y);    void solve();}AC;char str[maxn];int main () {    int cas = 0;    while (scanf("%s", str) == 1 && strcmp(str, "#")) {        AC.init(str);        AC.build(27);        AC.get_height();        printf("Case %d: ", ++cas);        AC.solve();    }    return 0;}void Suffix_Arr::init(char* str) {    n = 0;    int len = strlen(str);    for (int i = 0; i < len; i++)        s[n++] = str[i] - 'a' + 1;    s[n++] = 0;}void Suffix_Arr::solve() {    /*    for (int i = 0; i < n; i++)        printf("%d ", SA[i]);    printf("\n");    for (int i = 0; i < n; i++)        printf("%d ", height[i]);    printf("\n");    */    rmq_init();    int ans = 0;    vector<int> vec;    for (int l = 1; l < n; l++) {        for (int i = 0; i + l < n; i += l) {            int lcp = rmq_query(rank[i], rank[i + l]);            int k = lcp / l + 1;            int p = i - (l - lcp % l);            if (p >= 0 && lcp % l && rmq_query(rank[p], rank[p + l]) >= lcp)                k++;            if (k > ans) {                ans = k;                vec.clear();            }            if (k == ans)                vec.push_back(l);        }    }    int pos, len;    for (int i = 0; i < n; i++) {        bool flag = false;        for (int j = 0; j < vec.size(); j++) {            if (SA[i] + vec[j] >= n)                continue;            if (rmq_query(i, rank[SA[i] + vec[j]]) >= (ans - 1) * vec[j]) {                pos = SA[i];                len = vec[j] * ans;                flag = true;                break;            }        }        if (flag)            break;    }    for (int i = 0; i < len; i++)        printf("%c", s[pos + i] + 'a' - 1);    printf("\n");}void Suffix_Arr::rmq_init() {    for (int i = 0; i < n; i++) d[i][0] = height[i];    for (int k = 1; (1<<k) <= n; k++) {        for (int i = 0; i + (1<<k) - 1 < n; i++)            d[i][k] = min(d[i][k-1], d[i+(1<<(k-1))][k-1]);    }}int Suffix_Arr::rmq_query(int x, int y) {    if (x == y)        return d[x][0];    if (x > y)        swap(x, y);    x++;    int k = 0;    while ((1<<(k+1) <= y - x + 1)) k++;    return min(d[x][k], d[y - (1<<k) + 1][k]);}void Suffix_Arr::get_height() {    for (int i = 0; i < n; i++)        rank[SA[i]] = i;    int mv = height[n-1] = 0;    for (int i = 0; i < n - 1; i++) {        if (mv) mv--;        int j = SA[rank[i] - 1];        while (s[i+mv] == s[j+mv])            mv++;        height[rank[i]] = mv;    }}void Suffix_Arr::build (int m) {    int *x = tmp_one, *y = tmp_two;    for (int i = 0; i < m; i++) c[i] = 0;    for (int i = 0; i < n; i++) c[x[i] = s[i]]++;    for (int i = 1; i < m; i++) c[i] += c[i-1];    for (int i = n - 1; i >= 0; i--) SA[--c[x[i]]] = i;    for (int k = 1; k <= n; k <<= 1) {        int mv = 0;        for (int i = n - k; i < n; i++) y[mv++] = i;        for (int i = 0; i < n; i++) if (SA[i] >= k)            y[mv++] = SA[i] - k;        for (int i = 0; i < m; i++) c[i] = 0;        for (int i = 0; i < n; i++) c[x[y[i]]]++;        for (int i = 1; i < m; i++) c[i] += c[i-1];        for (int i = n - 1; i >= 0; i--) SA[--c[x[y[i]]]] = y[i];        swap(x, y);        mv = 1;        x[SA[0]] = 0;        for (int i = 1; i < n; i++)            x[SA[i]] = (y[SA[i-1]] == y[SA[i]] && y[SA[i-1] + k] == y[SA[i] + k] ? mv - 1 : mv++);        if (mv >= n)            break;        m = mv;    }}

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