UV 10829-L-Gap Substrings (suffix array)

UV 10829-L-Gap Substrings (suffix array)
Ultraviolet A 10829-L-Gap Substrings

Question Link

If a string is in the format of UGU and the length of G is L, the string is called L. If a string is given, ask the number of substrings of this string as g.

Idea: poj1_3 was created before this question. One idea is to enumerate the length segments, so that for a U with the length of l, as long as the lcp is located forward and backward after the current position (l + g) respectively, the length of the two lcp minus l is the number that can be used. The sum is the answer.

Code:

#include 
 
  #include 
  
   #include using namespace std;typedef long long ll;const int MAXLEN = 100005;struct Suffix {    int s[MAXLEN];    int sa[MAXLEN], t[MAXLEN], t2[MAXLEN], c[MAXLEN], n;    int rank[MAXLEN], height[MAXLEN];    int best[MAXLEN][20];    int g, len;    char str[MAXLEN];    void build_sa(int m) {n++;int i, *x = t, *y = t2;for (i = 0; i < m; i++) c[i] = 0;for (i = 0; i < n; i++) c[x[i] = s[i]]++;for (i = 1; i < m; i++) c[i] += c[i - 1];for (i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;for (int k = 1; k <= n; k <<= 1) {    int p = 0;    for (i = n - k; i < n; i++) y[p++] = i;    for (i = 0; i < n; i++) if (sa[i] >= k) y[p++] = sa[i] - k;    for (i = 0; i < m; i++) c[i] = 0;    for (i = 0; i < n; i++) c[x[y[i]]]++;    for (i = 0; i < m; i++) c[i] += c[i - 1];    for (i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];    swap(x, y);    p = 1; x[sa[0]] = 0;    for (i = 1; i < n; i++)x[sa[i]] = (y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k]) ? p - 1 : p++;    if (p >= n) break;    m = p;}n--;    }    void getHeight() {int i, j, k = 0;for (i = 1; i <= n; i++) rank[sa[i]] = i;for (i = 0; i < n; i++) {    if (k) k--;    int j = sa[rank[i] - 1];    while (s[i + k] == s[j + k]) k++;    height[rank[i]] = k;}    }    void initRMQ() {for (int i = 1; i <= n; i++) best[i][0] = height[i];for (int j = 1; (1<
   
     R) swap(L, R);L++;int k = 0;while ((1<<(k + 1)) <= R - L + 1) k++;return min(best[L][k], best[R - (1<
    
     = 0; i--)    s[n++] = str[i] - 'a' + 1;s[n] = 0;    }    ll solve() {init();build_sa(28);getHeight();initRMQ();ll ans = 0;for (int d = 1; d < len / 2; d++) {    for (int j = 0; j < len; j += d) {int l = j, r = j + d + g, sum = 0;if (r < len) sum += min(lcp(l, r), d);if (l >= 1) sum += min(lcp(n - l, n - r), d - 1);ans += max(0, sum - d + 1);    }}return ans;    }} gao;int t;int main() {    int cas = 0;    scanf("%d", &t);    while(t--) {printf("Case %d: %lld\n", ++cas, gao.solve());    }    return 0;}