【BZOJ3469】ADERA-优快云博客

本文介绍了一种将字典树与线段树相结合的数据结构解决方案，通过该方法可以高效地处理字符串查询问题。文章详细描述了如何构建这两种数据结构，并展示了如何在实际编程竞赛中使用它们来解决特定问题。

【题目链接】

点击打开链接

【思路要点】

字典树套线段树即可。
时空复杂度\(O(NLogN)\)。

【代码】

#includebitsstdc++.h
using namespace std;
const int MAXN = 1e5 + 5;
const int MAXP = 3e6 + 5;
template typename T void chkmax(T &x, T y) {x = max(x, y); }
template typename T void chkmin(T &x, T y) {x = min(x, y); } 
template typename T void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x  10 + c - '0';
	x = f;
}
template typename T void write(T x) {
	if (x  0) x = -x, putchar('-');
	if (x  9) write(x  10);
	putchar(x % 10 + '0');
}
template typename T void writeln(T x) {
	write(x);
	puts();
}
struct info {int val, home; };
char s[MAXN];
info tmp[MAXN];
int n, m, val[MAXN];
struct SegmentTree {
	struct Node {
		int lc, rc;
		int sum;
	} a[MAXP];
	int size, n;
	void init(int x) {
		n = x;
		size = 0;
	}
	void update(int root) {
		a[root].sum = a[a[root].lc].sum + a[a[root].rc].sum;
	}
	void modify(int &root, int l, int r, int val) {
		if (root == 0) root = ++size;
		if (l == r) {
			a[root].sum = 1;
			return;
		}
		int mid = (l + r)  2;
		if (mid = val) modify(a[root].lc, l, mid, val);
		else modify(a[root].rc, mid + 1, r, val);
		update(root);
	}
	void modify(int &root, int val) {
		modify(root, 0, n, val);
	}
	void query(int root, int l, int r, int cnt) {
		if (a[root].sum == 0) return;
		if (l == r) {
			putchar(' '), write(tmp[l].home);
			return;
		}
		int mid = (l + r)  2;
		if (a[a[root].lc].sum) query(a[root].lc, l, mid, min(cnt, a[a[root].lc].sum));
		cnt -= min(cnt, a[a[root].lc].sum);
		if (cnt != 0) query(a[root].rc, mid + 1, r, cnt);
	}
	void query(int root, int cnt) {
		int tmp = a[root].sum;
		write(tmp);
		chkmin(cnt, tmp);
		query(root, 0, n, cnt);
	}
} ST;
struct Trie {
	struct Node {
		int child[26];
		int root;
		vector int stuff;
	} a[MAXN];
	int root, size;
	void insert(char s, int from) {
		int now = root, len = strlen(s + 1);
		for (int i = 1; i = len; i++) {
			int tmp = s[i] - 'a';
			if (a[now].child[tmp] == 0) a[now].child[tmp] = ++size;
			now = a[now].child[tmp];
			a[now].stuff.push_back(from);
		}
	}
	void init() {
		for (int i = 1; i = size; i++)
		for (unsigned j = 0; j  a[i].stuff.size(); j++)
			ST.modify(a[i].root, val[a[i].stuff[j]]);
	}
	void query(int cnt, char s) {
		int now = root, len = strlen(s + 1);
		for (int i = 1; i = len; i++) {
			int tmp = s[i] - 'a';
			now = a[now].child[tmp];
			if (now == 0) break;
		}
		ST.query(a[now].root, cnt);
	}
} Trie;
bool cmp(info a, info b) {
	return a.val  b.val;
}
int main() {
	read(n), read(m);
	for (int i = 0; i = n - 1; i++) {
		read(val[i]);
		tmp[i] = (info) {val[i], i};
		int k; read(k);
		while (k--) {
			scanf( %s, s + 1);
			Trie.insert(s, i);
		}
	}
	sort(tmp, tmp + n, cmp);
	for (int i = 0; i = n - 1; i++)
		val[tmp[i].home] = i;
	ST.init(n - 1);
	Trie.init();
	while (m--) {
		int cnt; read(cnt);
		scanf( %s, s + 1);
		Trie.query(cnt, s);
		printf(n);
	}
	return 0;
}