后缀数组练习题若干

POJ 1743    不可重叠最长重复子串

二分答案。 即子串的长度，假设为k时。

利用height数组，将排序后的后缀分为若干组。

每组内的height值都不小于k。

然后只需查看组内是否有满足要求的两个不会产生重叠的子串即可。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <set>
#include <queue>
#include <algorithm>
#define MAXN 22222
#define MAXM 111
#define INF 1000000000
using namespace std;
int r[MAXN];
int wa[MAXN], wb[MAXN], wv[MAXN], tmp[MAXN];
int sa[MAXN]; //index range 1~n value range 0~n-1
int cmp(int *r, int a, int b, int l)
{
    return r[a] == r[b] && r[a + l] == r[b + l];
}
void da(int *r, int *sa, int n, int m)
{
    int i, j, p, *x = wa, *y = wb, *ws = tmp;
    for (i = 0; i < m; i++) ws[i] = 0;
    for (i = 0; i < n; i++) ws[x[i] = r[i]]++;
    for (i = 1; i < m; i++) ws[i] += ws[i - 1];
    for (i = n - 1; i >= 0; i--) sa[--ws[x[i]]] = i;
    for (j = 1, p = 1; p < n; j *= 2, 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]];
        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];
        for (swap(x, y), p = 1, x[sa[0]] = 0, i = 1; i < n; i++)
            x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;
    }
}
int rank[MAXN]; //index range 0~n-1 value range 1~n
int height[MAXN]; //index from 1   (height[1] = 0)
void calheight(int *r, int *sa, 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)
        for (k ? k-- : 0, j = sa[rank[i] - 1]; r[i + k] == r[j + k]; ++k);
    return;
}
int n, a[MAXN];
bool check(int mid, int n)
{
    int flag = 0;
    int mx = -1, mi = n;
    for(int i = 2; i <= n + 1; i++)
    {
        if((i == n + 1 && flag) || (height[i] < mid && flag))
        {
            flag = 0;
            mx = max(mx, sa[i - 1]);
            mi = min(mi, sa[i - 1]);
            if(mx - mi >= mid) return true;
            mi = n, mx = -1;
        }
        else if(height[i] >= mid)
        {
            flag = 1;
            mx = max(mx, sa[i - 1]);
            mi = min(mi, sa[i - 1]);
        }
    }
    return false;
}
int main()
{
    while(scanf("%d", &n) != EOF && n)
    {
        for(int i = 0; i < n; i++) scanf("%d", &a[i]);
        for(int i = 0; i < n - 1; i++) r[i] = a[i + 1] - a[i] + 89;
        r[--n] = 0;
        da(r, sa, n + 1, 200);
        calheight(r, sa, n);
        int low = 4, high = n / 2, ans = 0;
        while(low <= high)
        {
            int mid = (low + high) >> 1;
            if(check(mid, n))
            {
                low = mid + 1;
                ans = max(ans, mid);
            }
            else high = mid - 1;
        }
        if(ans < 4) printf("0\n");
        else printf("%d\n", ans + 1);
    }
    return 0;
}

 

 

POJ 3261 可重叠的出现K次的最长重复子串

还是二分子串长度。 后缀分为若干组，然后判断是否有一个组的size不小于k

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <set>
#include <queue>
#include <algorithm>
#define MAXN 22222
#define MAXM 111
#define INF 1000000000
using namespace std;
int r[MAXN];
int wa[MAXN], wb[MAXN], wv[MAXN], tmp[MAXN];
int sa[MAXN]; //index range 1~n value range 0~n-1
int cmp(int *r, int a, int b, int l)
{
    return r[a] == r[b] && r[a + l] == r[b + l];
}
void da(int *r, int *sa, int n, int m)
{
    int i, j, p, *x = wa, *y = wb, *ws = tmp;
    for (i = 0; i < m; i++) ws[i] = 0;
    for (i = 0; i < n; i++) ws[x[i] = r[i]]++;
    for (i = 1; i < m; i++) ws[i] += ws[i - 1];
    for (i = n - 1; i >= 0; i--) sa[--ws[x[i]]] = i;
    for (j = 1, p = 1; p < n; j *= 2, 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]];
        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];
        for (swap(x, y), p = 1, x[sa[0]] = 0, i = 1; i < n; i++)
            x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;
    }
}
int rank[MAXN]; //index range 0~n-1 value range 1~n
int height[MAXN]; //index from 1   (height[1] = 0)
void calheight(int *r, int *sa, 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)
        for (k ? k-- : 0, j = sa[rank[i] - 1]; r[i + k] == r[j + k]; ++k);
    return;
}
int n, k;
bool check(int mid)
{
    int cnt = 1;
    for(int i = 2; i <= n; i++)
    {
        if(height[i] < mid) cnt = 1;
        else cnt++;
        if(cnt >= k) return 1;
    }
    return 0;
}
int main()
{
    int m = 0;
    scanf("%d%d", &n, &k);
    for(int i = 0; i < n; i++)
    {
        scanf("%d", &r[i]);
        r[i]++;
        m = max(r[i], m);
    }
    r[n] = 0;
    da(r, sa, n + 1, m + 1);
    calheight(r, sa, n);
    int low = 1, high = n;
    int ans = 0;
    while(low <= high)
    {
        int mid = (low + high) >> 1;
        if(check(mid))
        {
            ans = max(ans, mid);
            low = mid + 1;
        }
        else high = mid - 1;
    }
    printf("%d\n", ans);
    return 0;
}

 

 

SPOJ SUBST1 求一个串中不同子串的个数

每个子串都是某个后缀的前缀

对于一个后缀。 它将产生n - sa[k]个前缀

但是有height[k]个前缀是跟前一个字符串的前缀相同。

故每个后缀的贡献是n - sa[k] - height[k]

求和即可

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <set>
#include <queue>
#include <algorithm>
#define MAXN 55555
#define MAXM 111
#define INF 1000000000
using namespace std;
int r[MAXN];
int wa[MAXN], wb[MAXN], wv[MAXN], tmp[MAXN];
int sa[MAXN]; //index range 1~n value range 0~n-1
int cmp(int *r, int a, int b, int l)
{
    return r[a] == r[b] && r[a + l] == r[b + l];
}
void da(int *r, int *sa, int n, int m)
{
    int i, j, p, *x = wa, *y = wb, *ws = tmp;
    for (i = 0; i < m; i++) ws[i] = 0;
    for (i = 0; i < n; i++) ws[x[i] = r[i]]++;
    for (i = 1; i < m; i++) ws[i] += ws[i - 1];
    for (i = n - 1; i >= 0; i--) sa[--ws[x[i]]] = i;
    for (j = 1, p = 1; p < n; j *= 2, 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]];
        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];
        for (swap(x, y), p = 1, x[sa[0]] = 0, i = 1; i < n; i++)
            x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;
    }
}
int rank[MAXN]; //index range 0~n-1 value range 1~n
int height[MAXN]; //index from 1   (height[1] = 0)
void calheight(int *r, int *sa, 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)
        for (k ? k-- : 0, j = sa[rank[i] - 1]; r[i + k] == r[j + k]; ++k);
    return;
}
char s[MAXN];
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%s", s);
        int n = strlen(s);
        int m = 0;
        for(int i = 0; i < n; i++)
        {
            r[i] = (int)s[i];
            m = max(m, r[i]);
        }
        r[n] = 0;
        da(r, sa, n + 1, m + 1);
        calheight(r, sa, n);
        long long ans = 0;
        for(int i = 1; i <= n; i++) ans += n - sa[i] - height[i];
        printf("%lld\n", ans);
    }
    return 0;
}

 

URAL 1297  求最长回文串

假设原串为S，将原串倒置后是T。

建立一个新串S+“~”+T

然后对新串做后缀数组。

然后我们枚举的是回文串的中心。

假设中心的位置为i。

有两种情况

回文为奇数

那么求lcp(i, n - i - 1)

回文为偶数那么求lcp(i, n - i)

然后更新最优解即可

用手画一画就知道是什么意思了。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <set>
#include <queue>
#include <algorithm>
#define MAXN 111111
#define MAXM 111
#define INF 1000000000
using namespace std;
int r[MAXN];
int wa[MAXN], wb[MAXN], wv[MAXN], tmp[MAXN];
int sa[MAXN]; //index range 1~n value range 0~n-1
int cmp(int *r, int a, int b, int l)
{
    return r[a] == r[b] && r[a + l] == r[b + l];
}
void da(int *r, int *sa, int n, int m)
{
    int i, j, p, *x = wa, *y = wb, *ws = tmp;
    for (i = 0; i < m; i++) ws[i] = 0;
    for (i = 0; i < n; i++) ws[x[i] = r[i]]++;
    for (i = 1; i < m; i++) ws[i] += ws[i - 1];
    for (i = n - 1; i >= 0; i--) sa[--ws[x[i]]] = i;
    for (j = 1, p = 1; p < n; j *= 2, 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]];
        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];
        for (swap(x, y), p = 1, x[sa[0]] = 0, i = 1; i < n; i++)
            x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;
    }
}
int rank[MAXN]; //index range 0~n-1 value range 1~n
int height[MAXN]; //index from 1   (height[1] = 0)
void calheight(int *r, int *sa, 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)
        for (k ? k-- : 0, j = sa[rank[i] - 1]; r[i + k] == r[j + k]; ++k);
    return;
}
int Log[MAXN];
int mi[MAXN][20];
void rmqinit(int n)
{
    for(int i = 1; i <= n; i++) mi[i][0] = height[i];
    int m = Log[n];
    for(int i = 1; i <= m; i++)
        for(int j = 1; j <= n; j++)
        {
            mi[j][i] = mi[j][i - 1];
            if(j + (1 << (i - 1)) <= n) mi[j][i] = min(mi[j][i], mi[j + (1 << (i - 1))][i - 1]);
        }
}
int lcp(int a, int b)
{
    a = rank[a];    b = rank[b];
    if(a > b) swap(a,b);
    a ++;
    int t = Log[b - a + 1];
    return min(mi[a][t] , mi[b - (1<<t) + 1][t]);
}
char s[MAXN * 2];
int main()
{
    Log[1] = 0;
    for(int i = 2; i < MAXN; i++) Log[i] = Log[i >> 1] + 1;
    while(scanf("%s", s) != EOF)
    {
        int len = strlen(s);
        for(int i = 0; i < len; i++) r[i] = (int)s[i];
        r[len] = 128;
        for(int i = 0; i < len; i++) r[len + 1 + i] = (int)s[len - 1 - i];
        int n = 2 * len + 1;
        r[n] = 0;
        da(r, sa, n + 1, 130);
        calheight(r, sa, n);
        rmqinit(n);
        int ans = 0;
        int pos;
        for(int i = 0; i < len; i++)
        {
            int tmp = lcp(i, n - i - 1); //奇数
            if(tmp * 2 - 1 > ans)
            {
                ans= tmp * 2 - 1;
                pos = i - tmp + 1;
            }
            tmp = lcp(i, n - i); //偶数
            if(tmp * 2 > ans)
            {
                ans = tmp * 2;
                pos = i - tmp;
            }
        }
        for(int i = 0; i < ans; i++) putchar(s[pos + i]);
        puts("");
    }
    return 0;
}

 

POJ 2406  

给定一个字符串S，已知该串是由某串重复K次 连接得到的。

求最大的k

这题的话。 貌似POJ上暴力跑的很快。

用后缀数组需要的求是枚举子串的长度。

假设长度为len， 那么检查lcp(0, len)是否等于n - len即可

倍增在这里被卡掉了

用的DC3

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <set>
#include <queue>
#include <cmath>
#include <algorithm>
#define MAXN 1111111
#define MAXM 111
#define INF 1000000000
#define F(x) ((x)/3+((x)%3==1?0:tb))
#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)
using namespace std;
int wa[MAXN] , wb[MAXN] , wv[MAXN] , tmp[MAXN];
int c0(int *r, int a, int b){
    return r[a] == r[b] && r[a + 1] == r[b + 1] && r[a + 2] == r[b + 2];
}
int c12(int k, int *r, int a, int b){
    if (k == 2)
    return r[a] < r[b] || r[a] == r[b] && c12(1, r, a + 1, b + 1);
    else return r[a] < r[b] || r[a] == r[b] && wv[a + 1] < wv[b + 1];
}
void sort(int *r, int *a, int *b, int n, int m)
{
    int i;
    for (i = 0; i < n; i++) wv[i] = r[a[i]];
    for (i = 0; i < m; i++) tmp[i] = 0;
    for (i = 0; i < n; i++) tmp[wv[i]]++;
    for (i = 1; i < m; i++) tmp[i] += tmp[i-1];
    for (i = n-1; i >= 0; i--) b[--tmp[wv[i]]] = a[i];
}
void dc3(int *r, int *sa, int n, int m)
{
    int i, j, *rn = r + n;
    int *san = sa + n, ta = 0, tb = (n + 1) / 3, tbc = 0, p;
    r[n] = r[n + 1] = 0;
    for (i = 0; i < n; i++) if (i % 3 != 0) wa[tbc++] = i;
    sort(r + 2, wa, wb, tbc, m);
    sort(r + 1, wb, wa, tbc, m);
    sort(r, wa, wb, tbc, m);
    for (p = 1, rn[F(wb[0])] = 0, i = 1; i < tbc; i++)
        rn[F(wb[i])] = c0(r, wb[i-1], wb[i]) ? p-1 : p++;
    if (p < tbc) dc3(rn, san, tbc, p);
    else for (i = 0; i < tbc; i++) san[rn[i]] = i;
    for (i = 0; i < tbc; i++) if (san[i] < tb) wb[ta++] = san[i] * 3;
    if (n % 3 == 1) wb[ta++] = n-1;
    sort(r, wb, wa, ta, m);
    for (i = 0; i < tbc; i++) wv[wb[i] = G(san[i])] = i;
    for (i = 0, j = 0, p = 0; i < ta && j < tbc; p++)
        sa[p] = c12(wb[j] % 3, r, wa[i], wb[j]) ? wa[i++] : wb[j++];
    for (; i < ta; p++) sa[p] = wa[i++];
    for (; j < tbc; p++) sa[p] = wb[j++];
}
void da(int str[], int sa[], int rank[], int height[], int n, int m)
{
//       for (int i = n; i < n * 3; i++)
//        str[i] = 0;
    dc3 (str , sa , n + 1 , m);
    int i, j, k;
    for (i = 0; i < n; i++){
        sa[i] = sa[i + 1];
        rank[sa[i]] = i;
    }
    for (i = 0, j = 0, k = 0; i < n; height[rank[i ++]] = k)
        if (rank[i] > 0)
            for (k ? k--: 0 , j = sa[rank[i]-1];
            i + k < n && j + k < n && str[i + k] == str[j + k];
            k++);
}
int lcp[MAXN];
int r[MAXN];
int  sa[MAXN], rank[MAXN] , height[MAXN];
int n;
void getlcp()
{
    int k = rank[0];
    lcp[k] = n;
    for(int i = k; i >= 2; i--)
        lcp[i - 1] = min(lcp[i], height[i]);
    for(int i = k + 1; i <= n; i++)
        lcp[i] = min(lcp[i - 1], height[i]);
}
char s[MAXN];
bool ok(int k)
{
    int rk = rank[k];
    if(lcp[rk] == n - k) return true;
    return false;
}
int main()
{
    while(gets(s))
    {
        if(s[0] == '.') break;
        n = strlen(s);
        for(int i = 0; i <= n; i++) r[i] = s[i];
        da(r, sa, rank, height, n + 1, 130);
        getlcp();
        int tmp = (int)sqrt(n + 0.5);
        int ans = 0;
        for(int i = 1; i <= tmp; i++)
        {
            if(n % i != 0) continue;
            if(ok(i)) ans = max(ans, n / i);
            if(ok(n / i)) ans = max(ans, i);
        }
        printf("%d\n", ans);
    }
    return 0;
}