G3. Good Substrings

来源：互联网发布：俄罗斯淘宝编辑：程序博客网时间：2024/04/30 01:32
http://codeforces.com/contest/316/problem/G3
利用RMQ 求后缀的前缀在每个串里面的个数
二分每个后缀满足要求的最大和最小前缀长度
我用了十个后缀数组
代码写得有些拖沓
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1typedef long long LL;const int INF = 0x7f7f7f7f;const int maxn = 111111;const int mod = 1000000000;const double eps = 1e-10;char s[maxn];char ss[12][maxn];int A[20], B[20];int Log[20];int a[maxn], wa[maxn], wb[maxn], wv[maxn], wn[maxn];int tp[maxn];struct range_min_query {    int i, j, k, r[18][maxn];    void make(int n, int arr[]) {        for (i = 0; i <= n; ++i)            r[0][i] = arr[i];        for (i = 0; i < 17; ++i) {            k = Log[i];            for (j = 0; j + k <= n; ++j)                r[i + 1][j] = min(r[i][j], r[i][j + k]);        }    }    int query(int L, int R) {        if (R == L)            return r[k][L];        k = tp[R - L];        return min(r[k][L], r[k][R - Log[k]]);    }} RMQ[10];struct suffix_array {    int sum[maxn], sa[maxn], rank[maxn], h[maxn];    bool 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 n, int m) {        int i, j, p, *t;        int *x = wa, *y = wb;        for (i = 0; i < m; ++i)            wn[i] = 0;        for (i = 0; i < n; ++i)            wn[x[i] = r[i]]++;        for (i = 1; i < m; ++i)            wn[i] += wn[i - 1];        for (i = n - 1; i >= 0; --i)            sa[--wn[x[i]]] = i;        for (j = 1, p = 1; p < n; j <<= 1, 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)                wn[i] = 0;            for (i = 0; i < n; ++i)                wn[wv[i]]++;            for (i = 1; i < m; ++i)                wn[i] += wn[i - 1];            for (i = n - 1; i >= 0; --i)                sa[--wn[wv[i]]] = y[i];            p = 1, t = x, x = y, y = t;            for (x[sa[0]] = 0, i = 1; i < n; ++i)                x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;        }    }    void Calheight(int r[], int n) {        int i, j, k = 0;        for (i = 1; i <= n; ++i)            rank[sa[i]] = i;        for (i = 0; i < n; h[rank[i++]] = k)            for (k ? k-- : 0, j = sa[rank[i] - 1]; r[i + k] == r[j + k]; ++k)                ;    }    int left(int l, int r, int k, range_min_query& RMQ) {        int m, p = r, t, ans = r;        while (l <= r) {            m = (l + r) >> 1;            t = RMQ.query(m + 1, p + 1);            if (t >= k)                ans = m, r = m - 1;            else                l = m + 1;        }        return ans;    }    int right(int l, int r, int k, range_min_query& RMQ) {        int m, p = l, t, ans = l;        while (l <= r) {            m = (l + r) >> 1;            t = RMQ.query(p + 1, m + 1);            if (t >= k)                ans = m, l = m + 1;            else                r = m - 1;        }        return ans;    }    int search(int p, int len, int k, range_min_query& RMQ) {        int L, R;        L = left(1, p, k, RMQ);        R = right(p, len, k, RMQ);        return getsum(L, R);    }    void Calsum(int n, int len) {        sum[0] = 0;        for (int i = 1; i <= n; ++i)            sum[i] = sum[i - 1] + (sa[i] > len && sa[i] < n);    }    int getsum(int L, int R) {        return sum[R] - sum[L - 1];    }} suffix[10], src;int tlen[15];bool low(int i, int n, int L) {    int l, r, tmp;    for (int k = 0; k < n; ++k) {        l = suffix[k].rank[i];        r = tlen[k];        tmp = suffix[k].search(l, r, L, RMQ[k]);        if (tmp > B[k])            return false;    }    return true;}bool high(int i, int n, int L) {    int l, r, tmp;    for (int k = 0; k < n; ++k) {        l = suffix[k].rank[i];        r = tlen[k];        tmp = suffix[k].search(l, r, L, RMQ[k]);        if (tmp < A[k])            return false;    }    return true;}int main() {    int i, len, j;    Log[0] = 1;    for (i = 1; i < 20; ++i)        Log[i] = Log[i - 1] << 1;    tp[1] = 0;    i = 0;    for (j = 2; j < maxn; ++j) {        if (Log[i + 1] < j)            tp[j] = ++i;        else            tp[j] = i;        //printf("%d %d\n",j,tp[j]);    }    scanf("%s", s);    len = strlen(s);    for (i = 0; i <= len; ++i)        a[i] = (int) s[i];    src.Da(a, len + 1, 128);    src.Calheight(a, len);    for (i = 0; i < 10; ++i)        for (j = 0; j < len; ++j)            ss[i][j] = s[j];    int n;    scanf("%d", &n);    for (i = 0; i < n; ++i) {        ss[i][len] = '$';        scanf("%s%d%d", ss[i] + len + 1, A + i, B + i);        tlen[i] = strlen(ss[i]);        for (j = 0; j <= tlen[i]; ++j)            a[j] = (int) ss[i][j];        suffix[i].Da(a, tlen[i] + 1, 128);        suffix[i].Calheight(a, tlen[i]);        suffix[i].Calsum(tlen[i], len);        RMQ[i].make(tlen[i] + 1, suffix[i].h);    }    int L, R, p, m, l, r;    int ans = 0;    for (i = 0; i < len; ++i) {        p = src.rank[i];        L = src.h[p] + 1;        R = len - i;        l = -1, r = -2;        while (L <= R) {            m = (L + R) >> 1;            if (low(i, n, m))                l = m, R = m - 1;            else                L = m + 1;        }        L = src.h[p] + 1;        R = len - i;        while (L <= R) {            m = (L + R) >> 1;            if (high(i, n, m))                r = m, L = m + 1;            else                R = m - 1;        }        if (r != -2 && l != -1 && r >= l)            ans += (r - l + 1);    }    printf("%d\n", ans);    return 0;}