題目鏈接:uva 10829 - L-Gap Substrings
題目大意:給定一個字符串,問有多少字符串滿足UVU的形式,要求U非空,V的長度為g。
解題思路;對字符串的正序和逆序構建後綴數組,然後枚舉U的長度l,每次以長度l分區間,在l和l+d+g所在的兩個區間上確定U的最大長度。
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int maxn = 100005;
struct Suffix_Arr {
int n, len, s[maxn];
int SA[maxn], rank[maxn], height[maxn];
int tmp_one[maxn], tmp_two[maxn], c[30];
int d[maxn][20];
void init(char* str);
void build_arr(int m);
void get_height();
void rmq_init();
int rmq_query(int l, int r);
ll solve(int m);
void put();
}AC;
char str[maxn];
int main () {
int cas, n;
scanf("%d", &cas);
for (int i = 1; i <= cas; i++) {
scanf("%d%s", &n, str);
AC.init(str);
printf("Case %d: %lld\n", i, AC.solve(n));
}
return 0;
}
void Suffix_Arr::put () {
for (int i = 0; i < n; i++)
printf("%d ", SA[i]);
printf("\n");
for (int i = 0; i < n; i++)
printf("%d ", height[i]);
printf("\n");
}
ll Suffix_Arr::solve(int m) {
build_arr(28);
get_height();
rmq_init();
ll ret = 0;
for (int l = 1; l < len / 2; l++) {
for (int i = 0; i < len; i += l) {
int j = i + l + m, sum = 0;
if (j < len)
sum += min(rmq_query(rank[i], rank[j]), l);
if (i)
sum += min(rmq_query(rank[n-i-1], rank[n-j-1]), l-1);
ret += max(0, sum - l + 1);
}
}
return ret;
}
int Suffix_Arr::rmq_query(int l, int r) {
if (l > r) swap(l, r);
l++;
int k = 0;
while ((1<<(k+1)) <= r - l + 1) k++;
return min(d[l][k], d[r - (1<= 0; i--)
s[n++] = str[i] - 'a' + 1;
s[n++] = 0;
}
void Suffix_Arr::get_height() {
for (int i = 0; i < n; i++)
rank[SA[i]] = i;
int mv = height[0] = 0;
for (int i = 0; i < n - 1; i++) {
if (mv) mv--;
int j = SA[rank[i] - 1];
while (s[i+mv] == s[j+mv])
mv++;
height[rank[i]] = mv;
}
}
void Suffix_Arr::build_arr(int m) {
int *x = tmp_one, *y = tmp_two;
for (int i = 0; i < m; i++) c[i] = 0;
for (int i = 0; i < n; i++) c[x[i] = s[i]]++;
for (int i = 1; i < m; i++) c[i] += c[i-1];
for (int i = n-1; i >= 0; i--) SA[--c[x[i]]] = i;
for (int k = 1; k <= n; k <<= 1) {
int mv = 0;
for (int i = n - k; i < n; i++) y[mv++] = i;
for (int i = 0; i < n; i++) if (SA[i] >= k)
y[mv++] = SA[i] - k;
for (int i = 0; i < m; i++) c[i] = 0;
for (int i = 0; i < n; i++) c[x[y[i]]]++;
for (int i = 1; i < m; i++) c[i] += c[i-1];
for (int i = n-1; i >= 0; i--) SA[--c[x[y[i]]]] = y[i];
swap(x, y);
mv = 1;
x[SA[0]] = 0;
for (int i = 1; i < n; i++)
x[SA[i]] = (y[SA[i-1]] == y[SA[i]] && y[SA[i-1] + k] == y[SA[i] + k] ? mv - 1 : mv++);
if (mv >= n)
break;
m = mv;
}
}