hdu Boring counting

Boring counting

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 120 Accepted Submission(s): 52

Problem Description

035 now faced a tough problem,his english teacher gives him a string,which consists with n lower case letter,he must figure out how many substrings appear at least twice,moreover,such apearances can not overlap each other. Take aaaa as an example.”a” apears four times,”aa” apears two times without overlaping.however,aaa can’t apear more than one time without overlaping.since we can get “aaa” from [0-2](The position of string begins with 0) and [1-3]. But the interval [0-2] and [1-3] overlaps each other.So “aaa” can not take into account.Therefore,the answer is 2(“a”,and “aa”).

Input

The input data consist with several test cases.The input ends with a line “#”.each test case contain a string consists with lower letter,the length n won’t exceed 1000(n <= 1000).

Output

For each test case output an integer ans,which represent the answer for the test case.you’d better use int64 to avoid unnecessary trouble.

Sample Input

aaaa
ababcabb
aaaaaa
#

Sample Output

2
3
3

Source

2010 ACM-ICPC Multi-University Training Contest（9）——Host by HNU

Recommend

zhengfeng

分析：倍增算法写的后缀数组。因为要求重复的子串数量，枚举子串长度，利用公共前缀height数组确定是否可行。

#include<cstdio>
#include<cstring>
#define MAXN 500010
#define maxn 1010
int ws[MAXN], wv[MAXN], wa[MAXN], wb[MAXN], sa[MAXN], rank[MAXN];
int num[maxn], height[maxn];
char str[maxn];
int m;

int min(int a, int b) {
    return a < b ? a : b;
}

int max(int a, int b) {
    return a > b ? a : b;
}

void make_sa(int n) {
    int i, j, *x, *y, *t, p;
    x = wa;
    y = wb;
    for (i = 0; i < m; ++i)
        ws[i] = 0;
    for (i = 0; i < n; ++i)
        ws[x[i] = num[i]]++;
    for (i = 1; i < m; ++i)
        ws[i] += ws[i - 1];
    for (i = n - 1; i >= 0; --i)
        sa[--ws[x[i]]] = i;
    p = 1;
    for (j = 1; p < n; j *= 2) {
        p = 0;
        for (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]];
        }
        for (i = 0; i < m; ++i)
            ws[i] = 0;
        for (i = 0; i < n; ++i)
            ws[wv[i]]++;
        for (i = 1; i < m; ++i)
            ws[i] += ws[i - 1];
        for (i = n - 1; i >= 0; --i)
            sa[--ws[wv[i]]] = y[i];
        t = x;
        x = y;
        y = t;
        p = 1;
        x[sa[0]] = 0;
        for (i = 1; i < n; ++i) {
            if (y[sa[i]] == y[sa[i - 1]] && y[sa[i] + j] == y[sa[i - 1] + j])
                x[sa[i]] = p - 1;
            else
                x[sa[i]] = p++;
        }
        m = p;
    }
}

void make_height(int n) {
    int i, j, k = 0;
    for (i = 1; i <= n; ++i)
        rank[sa[i]] = i;
    for (i = 0; i < n; ++i) {
        if (k)
            --k;
        j = sa[rank[i] - 1];
        while (num[i + k] == num[j + k])
            ++k;
        height[rank[i]] = k;
    }
}

int tot(int len, int n) {
    int i, mmin, mmax, cnt;
    mmin = 1 << 30;
    mmax = -1;
    cnt = 0;
    for (i = 2; i <= n; ++i) {
        //    printf("height=%d len=%d\n", height[i], len);
        if (height[i] >= len) {
            mmax = max(max(sa[i], sa[i - 1]), mmax);
            mmin = min(min(sa[i], sa[i - 1]), mmin);
            //      printf("mm %d %d\n", mmin, mmax);
        } else {
            if (mmax - mmin >= len) {
                ++cnt;
            }
            mmin = 1 << 30;
            mmax = -1;
        }
    }
    if (mmax - mmin >= len) {
        ++cnt;
    }
    return cnt;
}

int main() {
    int i, n, sum;
    while (scanf("%s", str), str[0] != '#') {
        sum = 0;
        m = 30;
        n = strlen(str);
        for (i = 0; i < n; ++i)
            num[i] = str[i] - 'a' + 1;
        num[n] = 0;
        make_sa(n + 1);
        make_height(n);
        for (i = 1; i <= n / 2; ++i)
            sum += tot(i, n);
        printf("%d\n", sum);
    }
    return 0;
}