Codeforces Round #189 (Div. 1) D. Have You Ever Heard About the Word?

 

当前找出所有最短的重复串，删去之后，不会再出现小于等于当前长度的重复串了。

那么重复串的长度最多有 $O(\sqrt n)$ 种，删去就用后缀数组实现，枚举当前长度的分割点，求公共前缀长度和公共后缀长度，就是当前重复的长度了，然后就打标记删去即可。

#include 

const int N = 1e5 + 7;

namespace SA {
    int sa[N], rk[N], fir[N], sec[N], c[N], height[N];
    char s[N]; // 有时候可以用int数组防止爆char
    int mn[N][20], lg[N];
    inline int MIN(int a, int b) {
        return height[a] < height[b] ? a : b;
    }
    void init(int len, char *str, int num = 122) {
        register int i, j, k;
        s[len + 1] = 0;
        for (i = 1; i <= len; i++) s[i] = s[2 * len + 1 - i + 1] = str[i];
        len = 2 * len + 1;
        s[len + 1] = 0;
        for (i = 1; i <= num; i++) c[i] = 0;
        for (i = 1; i <= len; i++) ++c[fir[i] = s[i]];
        for (i = 1; i <= num; i++) c[i] += c[i - 1];
        for (i = len; i >= 1; i--) sa[c[fir[i]]--] = i;
        for (k = 1; k <= len; k <<= 1) {
            int cnt = 0;
            for (i = len - k + 1; i <= len; i++) sec[++cnt] = i;
            for (i = 1; i <= len; i++) if (sa[i] > k) sec[++cnt] = sa[i] - k;
            for (i = 1; i <= num; i++) c[i] = 0;
            for (i = 1; i <= len; i++) ++c[fir[i]];
            for (i = 1; i <= num; i++) c[i] += c[i - 1];
            for (i = len; i >= 1; i--) sa[c[fir[sec[i]]]--] = sec[i], sec[i] = 0;
            std::swap(fir, sec);
            fir[sa[1]] = 1; cnt = 1;
            for (i = 2; i <= len; i++)
                fir[sa[i]] = (sec[sa[i]] == sec[sa[i - 1]] && sec[sa[i] + k] == sec[sa[i - 1] + k]) ? cnt : ++cnt;
            if (cnt == len) break;
            num = cnt;        
        }
        k = 0;
        for (i = 1; i <= len; i++) rk[sa[i]] = i;
        for (i = 1; i <= len; i++) {
            if (rk[i] == 1) continue;
            if (k) k--;
            j = sa[rk[i] - 1];
            while (j + k <= len && i + k <= len && s[i + k] == s[j + k]) k++;
            height[rk[i]] = k;
        }
        for (i = 1, j = 0; i <= len; i++) {
            lg[i] = (1 << (j + 1)) == i ? ++j : j;
            mn[i][0] = i;
        }
        for (int j = 1; j < 20; j++)
            for (int i = 1; i + (1 << j) - 1 <= len; i++)
                mn[i][j] = MIN(mn[i][j - 1], mn[i + (1 << (j - 1))][j - 1]);
    }
    inline int RMQ(int a, int b) {
        int log = lg[b - a + 1];
        b -= (1 << log) - 1;
        return MIN(mn[a][log], mn[b][log]);
    }
    inline int LCP(int i, int j) {
        i = rk[i], j = rk[j];
        if (i > j) std::swap(i, j);
        return height[RMQ(i + 1, j)];
    }
}

int cur, n, tag[N];
char s[N];

bool solve() {
    while (++cur <= n) {
        int last = 0;
        for (int i = cur, j = 2 * cur; j <= n; i += cur, j += cur) {
            int pre = SA::LCP(2 * n + 2 - i, 2 * n + 2 - j);
            int suf = SA::LCP(i, j);
            if (pre > cur) pre = cur;
            if (suf > cur) suf = cur;
            if (pre + suf - 1 >= cur) {
                int from = i - pre + 1, to = i + suf - cur;
                if (to > last) {
                    int pos = std::max(last + 1, from);
                    last = pos + cur - 1;
                    tag[last] = cur;
                }
            }
        }
        if (last) {
            for (int i = n; i >= 1; i--)
                if (tag[i] >= 2) tag[i - 1] = tag[i] - 1;
            int curn = 0;
            for (int i = 1; i <= n; i++)
                if (!tag[i]) s[++curn] = s[i];
                else tag[i] = 0;
            n = curn;
            s[n + 1] = 0;
            return 1;
        }
    }
    return 0;
}

int main() {
    scanf("%s", s + 1);
    n = strlen(s + 1);
    SA::init(n, s);
    while (solve()) SA::init(n, s);
    puts(s + 1);
    return 0;
}

View Code