lightoj 1428(后缀数组)

题意：给出字符串A和字符串B，问A的多少个不同子串内不含B。

题解：这个是看题解的，多加了一个rmax数组，rmax[i]表示A串的第i个位置(从0开始)最多向后延伸多少个字符的子串可以不包含B，相当于给出A的每个后缀的右范围，rmax数组的确定可以先把A和B串用一个很大的字符间隔并连接，然后得到height[i]，sa[i]，rank[i]，在按字典序排好的后缀串中，从rank[lenA+1] + 1的位置开始的串一定比B的字典序大，所以会包含B串，那么如果height[i] >= lenB，那么rmax[sa[i]] = lenB - 1，表示从在A串中sa[i]这个位置只能向后延伸lenB - 1，否则就会包含B，然后从lenA+lenB向前，如果rmax[i]没有被赋值，rmax[i] = rmax[i + 1] + 1，从而确定所有的rmax[i]，A串重新求height，sa，res += min(lenA - sa[i]，rmax[sa[i]]) - height[i]。

#include <cstdio>
#include <cstring>
#include <algorithm>
#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;
const int N = 100005;
int wa
, wb
, ws
, wv
, sa[N * 3];
int rank[N * 3], height[N * 3], s
, f[N * 3][35];
char str1
, str2
;
int rmax
;

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);
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) {
for (int i = 0; i < n; i++) wv[i] = r[a[i]];
for (int i = 0; i < m; i++) ws[i] = 0;
for (int i = 0; i < n; i++) ws[wv[i]]++;
for (int i = 1; i < m; i++) ws[i] += ws[i - 1];
for (int i = n - 1; i >= 0; i--) b[--ws[wv[i]]] = a[i];
}

void dc3(int *r, int *sa, int n, int m) {
int i, j, *rn = r + n, *san = sa + n, ta = 0, tb = (n + 1) / 3, tbc = 0, p;
r
= 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 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++);
}

int main() {
int t, cas = 1;
scanf("%d", &t);
while (t--) {
scanf("%s%s", str1, str2);
int len1 = strlen(str1), len2 = strlen(str2);
int cnt = 0;
for (int i = 0; i < len1; i++)
s[cnt++] = str1[i] - 'a' + 1;
s[cnt++] = 30;
for (int i = 0; i < len2; i++)
s[cnt++] = str2[i] - 'a' + 1;
s[cnt] = 0;
dc3(s, sa, cnt + 1, 50);
calheight(s, sa, cnt);

memset(rmax, -1, sizeof(rmax));
for (int i = rank[len1 + 1] + 1; i < cnt; i++)
if (height[i] >= len2)
rmax[sa[i]] = len2 - 1;
else break;
for (int i = cnt - 1; i >= 0; i--)
if (rmax[i] < 0)
rmax[i] = rmax[i + 1] + 1;

s[len1] = 0;
dc3(s, sa, len1 + 1, 50);
calheight(s, sa, len1);
int res = 0;
for (int i = 1; i <= len1; i++) {
int temp = min(rmax[sa[i]], len1 - sa[i]);
if (temp - height[i] > 0)
res += temp - height[i];
}
printf("Case %d: %d\n", cas++, res);
}
return 0;
}