【BZOJ2754】[SCOI2012]喵星球上的点名
【BZOJ2754】[SCOI2012]喵星球上的点名
题面
题解
这题有各种神仙做法啊，什么暴力\(AC\)自动机、\(SAM\)等等五花八门
我这个蒟蒻在这里提供一种复杂度正确且常数小的做法。
根据后缀数组经典套路，
我们用一个未出现过的字符将所有串连接起来求一边\(SA\)（不算询问串）
然后因为我们现在已经将所有后缀排好序了
而询问串要有贡献一定是一个后缀的前缀
所以一个区间能产生贡献的后缀是排名连续的一段
这样这段区间的左右端点可以二分出来
我们再对于每个位置记一下它是哪一个人的
这样第一问就被转化成了区间内有多少种颜色，直接上莫队在排名上跑即可
到了这一步
第二问就很好做了
我们考虑差分，每新遇到一类数，我们加上剩余询问个数的贡献
每去掉一类数，减去剩余询问个数的贡献即可
代码
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
inline int gi() {
register int data = 0, w = 1;
register char ch = 0;
while (!isdigit(ch) && ch != '-') ch = getchar();
if (ch == '-') w = -1, ch = getchar();
while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();
return w * data;
}
const int MAX_N = 2e5 + 5;
int N, n, m, a[MAX_N], id[MAX_N], lim = 1e4 + 5;
int sa[MAX_N];
void GetSA() {
#define cmp(i, j, k) (y[i] == y[j] && y[i + k] == y[j + k])
static int x[MAX_N], y[MAX_N], bln[MAX_N];
int M = lim;
for (int i = 1; i <= N; i++) bln[x[i] = a[i]]++;
for (int i = 1; i <= M; i++) bln[i] += bln[i - 1];
for (int i = N; i >= 1; i--) sa[bln[x[i]]--] = i;
for (int k = 1; k <= N; k <<= 1) {
int p = 0;
for (int i = 0; i <= M; i++) y[i] = 0;
for (int i = N - k + 1; i <= N; i++) y[++p] = i;
for (int i = 1; i <= N; i++) if (sa[i] > k) y[++p] = sa[i] - k;
for (int i = 0; i <= M; i++) bln[i] = 0;
for (int i = 1; i <= N; i++) bln[x[y[i]]]++;
for (int i = 1; i <= M; i++) bln[i] += bln[i - 1];
for (int i = N; i >= 1; i--) sa[bln[x[y[i]]]--] = y[i];
swap(x, y); x[sa[1]] = p = 1;
for (int i = 2; i <= N; i++) x[sa[i]] = cmp(sa[i], sa[i - 1], k) ? p : ++p;
if (p >= N) break;
M = p;
}
}
const int LEN = 400;
int cnt[MAX_N], bln[MAX_N];
struct Query { int l, r, id; } q[MAX_N]; int q_cnt = 0;
inline bool operator < (const Query &a, const Query &b) {
if (bln[a.l] ^ bln[b.l]) return a.r < b.r;
else return (bln[a.l] & 1) ? a.r < b.r : a.r > b.r;
}
int A1, ans1[MAX_N], ans2[MAX_N];
void add(int x, int pos) {
cnt[id[sa[x]]]++;
if (cnt[id[sa[x]]] == 1) ++A1, ans2[id[sa[x]]] += q_cnt - pos + 1;
}
void del(int x, int pos) {
cnt[id[sa[x]]]--;
if (cnt[id[sa[x]]] == 0) --A1, ans2[id[sa[x]]] -= q_cnt - pos + 1;
}
int main () {
#ifndef ONLINE_JUDGE
freopen("cpp.in", "r", stdin);
#endif
n = gi(), m = gi();
for (int i = 1, L; i <= n; i++) {
L = gi(); for (int j = 1; j <= L; j++) a[++N] = gi(), id[N] = i; a[++N] = ++lim;
L = gi(); for (int j = 1; j <= L; j++) a[++N] = gi(), id[N] = i; a[++N] = ++lim;
}
GetSA();
for (int i = 1; i <= m; i++) {
int L = 1, R = N;
for (int len = gi(), j = 1; j <= len; ++j) {
int x = gi(), l = L, r = R;
while (l <= r) {
int mid = (l + r) >> 1;
if (a[sa[mid] + j - 1] < x) l = mid + 1;
else r = mid - 1;
}
int tmp = l; l = L, r = R;
while (l <= r) {
int mid = (l + r) >> 1;
if (a[sa[mid] + j - 1] <= x) l = mid + 1;
else r = mid - 1;
}
L = tmp, R = r;
}
if (L <= R) q[++q_cnt] = (Query){L, R, i};
}
for (int i = 1; i <= N; i++) bln[i] = (i - 1) / LEN + 1;
sort(&q[1], &q[q_cnt + 1]);
for (int ql = 1, qr = 0, i = 1; i <= q_cnt; i++) {
while (ql < q[i].l) del(ql, i), ++ql;
while (ql > q[i].l) --ql, add(ql, i);
while (qr < q[i].r) ++qr, add(qr, i);
while (qr > q[i].r) del(qr, i), --qr;
ans1[q[i].id] = A1;
}
for (int i = 1; i <= m; i++) printf("%d\n", ans1[i]);
for (int i = 1; i <= n; i++) printf("%d ", ans2[i]);
return 0;
}
跑得还是挺快的，在洛谷跑了\(Rank4\)
