十二省联考2019 字符串问题
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原题链接
不会( ext{SAM})的蒟蒻只好用( ext{SA})来水了。
先读懂题意(窝在读题意这儿花了快(20)min).
简单来说,是给你一个字符串,然后给你两个区间的集合(A,B),一共有(m)组(A_i)到(B_j)的支配关系。若(A_i,B_j)有支配关系,则从(A_i)到(B_j)连边;若(B_j)是(A_i)的前缀,则珂以从(B_j)连向(A_i)。每个(A_i)有权值,是(A_i)区间的长度。当你把上述的图建好后,她要你求图中的最长路。如果珂以无限长,输出-1
。
一看就懵逼的神仙字符串图论题
来肝这题的神仙们应该都知道最长路咋求吧?就是先拓扑排序一下,然后从后往前( ext{dp})。这里就不讲了。
分析一下,这张图上一共的点数是不超过(|A|+|B|),即是线性的。但边数珂能高达(m+|A|cdot|B|)。再仔细地分析一下,从(A_i)到(B_j)连的边数是(m),是线性的。问题就转化成了优化(B_j)向(A_i)连边的过程。
结合(B_j)是(A_i)的前缀,我们很容易想到用( ext{SA})。我们找到(B_j)在( ext{sa})上的位置,设为( ext{pos})。(B_j)连向的所有(A_i)必须满足( ext{lcp})是(ge|B_j|)的,在( ext{sa})上必须是一个区间,且包含( ext{pos})。设这个区间为([l,r]),那么必须满足:(minlimits_{l+1le ile r}height_ige |B_j|),且(l)尽量忘左,(r)尽量往右。找(l,r)的过程珂以通过在( ext{height})数组上建出( ext{ST})表之后二分求出。然后只要再用线段树优化建图就珂以了。
但这样有个问题:你不能保证在([l,r])区间内的(|A_i|ge |B_j|)。这样只能得到(80)分。
所以我们需要考虑更好的做法:用可持久化线段树优化建图。按照(|A_i|)从大到小建树,每个(A_i)只在第(n-|A_i|)棵可持久化线段树建树时加入。在可持久化线段树上,第(i)棵树上的点要向第(i-1)棵树上的对应点连边(除非没有),这样珂以保证每一次加入的(B_j)的长度(le |A_i|)。具体细节就康代码吧。
总时间复杂度:(O(Nlog N)),空间复杂度:(O(Nlog N))
代码:
// Code by H~$~C
#include <bits/stdc++.h>
using namespace std;
#ifndef LOCAL_JUDGE
static char _in_buf[100000], *_in_p1 = _in_buf, *_in_p2 = _in_buf;
#define gc (__builtin_expect(_in_p1 == _in_p2, 0) && (_in_p2 = (_in_p1 = _in_buf) + fread(_in_buf, 1, 100000, stdin), _in_p1 == _in_p2) ? -1 : *_in_p1++)
#else
#define gc getchar()
#endif
inline int read() {
register char ch = gc;
register int x = 0;
while (ch < 48 || ch > 57) ch = gc;
while (ch > 47 && ch < 58) x = (x << 3) + (x << 1) + (ch ^ 48), ch = gc;
return x;
}
static const int Maxn = 200005;
static const int Maxs = 5000005;
int n, na, nb, N, m;
char str[Maxn];
int la[Maxn], ra[Maxn], lb[Maxn], rb[Maxn];
vector<int> A[Maxn];
int ST[Maxn][20];
int wa[Maxn], wb[Maxn], wc[Maxn], _s[Maxn];
int sa[Maxn], rnk[Maxn], height[Maxn];
template<typename T>
void build_SA(T *ss, int n, int m) {
register int *x = wa, *y = wb, *tmp;
register int i, j, w;
for (i = 1; i <= n; ++i) _s[i] = ss[i];
for (i = 1; i <= n; ++i) x[i] = _s[i], y[i] = i;
for (i = 1; i <= m; ++i) wc[i] = 0;
for (i = 1; i <= n; ++i) wc[x[i]]++;
for (i = 2; i <= m; ++i) wc[i] += wc[i - 1];
for (i = n; i >= 1; --i) sa[wc[x[y[i]]]--] = y[i];
for (w = 1; w <= n; w <<= 1) {
register int tot = 0;
for (i = n - w + 1; i <= n; ++i) y[++tot] = i;
for (i = 1; i <= n; ++i) if (sa[i] > w) y[++tot] = sa[i] - w;
for (i = 1; i <= m; ++i) wc[i] = 0;
for (i = 1; i <= n; ++i) wc[x[i]]++;
for (i = 2; i <= m; ++i) wc[i] += wc[i - 1];
for (i = n; i >= 1; --i) sa[wc[x[y[i]]]--] = y[i];
tmp = x, x = y, y = tmp, x[sa[1]] = 1, tot = 1;
for (i = 2; i <= n; ++i)
x[sa[i]] = ((y[sa[i]] == y[sa[i - 1]] && y[sa[i] + w] == y[sa[i - 1] + w]) ? tot : ++tot);
if (tot == n) break; m = tot;
}
for (i = 1; i <= n; ++i) rnk[sa[i]] = i;
for (i = 1, w = 0; i <= n; ++i) {
if (rnk[i] == 1) continue;
if (w) --w;
int j = sa[rnk[i] - 1];
while (i + w <= n && j + w <= n && _s[i + w] == _s[j + w]) w++;
height[rnk[i]] = w;
}
}
// sa[rank] = name, rnk[name] = rank
// height[i] = lcp(suffix(sa[i - 1]), suffix(sa[i]))
vector<int> g[Maxs];
int deg[Maxs], val[Maxs];
inline void clear_all(int u) { g[u].clear(), deg[u] = val[u] = 0; }
inline void add_edge(int u, int v) { g[u].push_back(v), deg[v]++; }
struct Node {
int id;
Node *l, *r;
Node() { }
Node(int id, Node *l, Node *r)
: id(id), l(l), r(r) { }
} *root[Maxn], pool[Maxs], *cur_pointer = pool;
inline Node *newnode(int id, Node *l = NULL, Node *r = NULL) {
return &(*++cur_pointer = Node(id, l, r));
}
int insert(Node *&p, int l, int r, int pos) {
if (!p) {
p = newnode(++N);
clear_all(N);
}
else {
p = newnode(p->id, p->l, p->r);
clear_all(++N);
add_edge(N, p->id);
p->id = N;
}
if (l == r) return p->id;
int mid = (l + r) >> 1, res;
if (pos <= mid) res = insert(p->l, l, mid, pos);
else res = insert(p->r, mid + 1, r, pos);
if (p->l) add_edge(p->id, p->l->id);
if (p->r) add_edge(p->id, p->r->id);
return res;
}
void link_edge(Node *p, int l, int r, int L, int R, int u) {
if (!p) return ;
if (L == l && r == R) return add_edge(u, p->id);
int mid = (l + r) >> 1;
if (R <= mid) return link_edge(p->l, l, mid, L, R, u);
if (L > mid) return link_edge(p->r, mid + 1, r, L, R, u);
link_edge(p->l, l, mid, L, mid, u);
link_edge(p->r, mid + 1, r, mid + 1, R, u);
}
int que[Maxs], qh, qe;
long long dp[Maxs];
void solve() {
cur_pointer = pool;
register int i, j;
register char ch;
while ((ch = gc) < 33);
for (n = 0; ch > 32; ch = gc) str[++n] = ch;
for (na = read(), i = 1; i <= na; ++i)
la[i] = read(), ra[i] = read();
for (nb = read(), i = 1; i <= nb; ++i)
lb[i] = read(), rb[i] = read();
N = na + nb;
build_SA<char>(str, n, 128);
for (i = 1; i <= n; ++i) ST[i][0] = height[i];
for (j = 1; j < 20; ++j)
for (i = 1; i + (1 << j) - 1 <= n; ++i)
ST[i][j] = min(ST[i][j - 1], ST[i + (1 << (j - 1))][j - 1]);
for (i = 1; i <= N; ++i) clear_all(i);
for (m = read(), i = 1; i <= m; ++i) {
int x = read(), y = read();
add_edge(x, y + na);
}
for (i = 1; i <= n; ++i) A[i].clear();
for (i = 1; i <= na; ++i) {
val[i] = ra[i] - la[i] + 1;
A[val[i]].push_back(i);
}
root[n + 1] = NULL;
for (i = n; i >= 1; --i) {
root[i] = root[i + 1];
for (int &x: A[i]) {
int id = insert(root[i], 1, n, rnk[la[x]]);
add_edge(id, x);
}
}
for (i = 1; i <= nb; ++i) {
int l = rnk[lb[i]], r = rnk[lb[i]];
int len = rb[i] - lb[i] + 1;
/* get the leftest position */ {
for (j = 0; l - (1 << j) + 1 >= 2; ++j);
for (; ~--j; ) {
if (l - (1 << j) + 1 >= 2 && ST[l - (1 << j) + 1][j] >= len) {
l -= (1 << j);
}
}
}
/* get the rightest position */ {
for (j = 0; r + (1 << j) <= n; ++j);
for (; ~--j; ) {
if (r + (1 << j) <= n && ST[r + 1][j] >= len) {
r += (1 << j);
}
}
}
link_edge(root[len], 1, n, l, r, i + na);
}
qh = qe = 0;
for (i = 1; i <= N; ++i) {
if (!deg[i]) que[qe++] = i;
}
while (qh < qe) {
int u = que[qh++];
for (int &v: g[u]) {
if (!--deg[v]) {
que[qe++] = v;
}
}
}
if (qe != N) {
puts("-1");
return ;
}
for (i = N - 1; ~i; --i) {
int u = que[i];
dp[u] = 0;
for (int &v: g[u]) {
dp[u] = max(dp[u], dp[v]);
}
dp[u] += val[u];
}
printf("%lld
", *max_element(dp + 1, dp + N + 1));
}
int main() {
int tests = read();
while (tests--) solve();
return 0;
}
跑得炒鸡慢(ldots)最大的点用了(7.40)秒
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