ural 1542 字典树（思维）

题目：http://acm.timus.ru/problem.aspx?space=1&num=1542

题意：

给出n个单词和各自出现的频率，然后给出m个某些单词的开头，要求找出频率最高不超过10个的以这些字符串开头的单词。

分析：

显然是字典树的题目，如果把单词建字典树，然后对于每个开头字符串，都查询一遍字典树，显然会超时。

所以逆向思考一下，把开头字符串建立字典树，然后查询每个单词，每个开头字符串找出不超过10个单词，这样就很容易实现了~~

#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
#include <queue>
#include <set>

using namespace std;
typedef long long ll;
typedef pair<int, int>pii;
const double PI = acos (-1.0);
const int INF = 0x3f3f3f3f;
const int N = 3e5 + 9;

int to
, cnt
;
char ans
[11][16], s[16];
struct Item {
char arr[16];
int num;
bool operator < (const Item& rhs) const {
if (num == rhs.num) return strcmp (arr, rhs.arr) < 0 ? 1 : 0;
return num > rhs.num;
}
} p
;
const int maxnode = 3e5 + 10;
struct Trie {
int ch[maxnode][26];
int val[maxnode];
int sz;
void clear() {
sz = 1;
memset (val, 0, sizeof (val) );
memset (ch[0], 0, sizeof (ch[0]) );
}
int idx (char c) {
return c - 'a';
}

void insert (const char *s, int id) {
int u = 0, n = strlen (s);
for (int i = 0; i < n; i++) {
int c = idx (s[i]);
if (!ch[u][c]) {
memset (ch[sz], 0, sizeof (ch[sz]) );
ch[u][c] = sz++;
}
u = ch[u][c];
}
if (!val[u]) val[u] = id, to[id] = id;
else to[id] = val[u];
}
void search (const char *s) {
int u = 0, len = strlen (s);
for (int i = 0; i < len; i++) {
int c = idx (s[i]);
u = ch[u][c];
if (u) {
if (val[u] && cnt[val[u]] <= 9)
strcpy (ans[val[u]][cnt[val[u]]++], s);
} else break;
}
}
};
Trie trie;
int main() {
//freopen ("f.txt", "r", stdin);
int n, m;
trie.clear();
memset (cnt, 0, sizeof (cnt) );
scanf ("%d", &n);
for (int i = 1; i <= n; i++) scanf ("%s%d", p[i].arr, &p[i].num);
sort (p + 1, p + 1 + n);
scanf ("%d", &m);
for (int i = 1; i <= m; i++) {
scanf ("%s", s);
trie.insert (s, i);
}
for (int i = 1; i <= n; i++) {
trie.search (p[i].arr);
}
bool flag = 0;
for (int i = 1; i <= m; i++) {
if (flag) printf ("\n");
flag = 1;
for (int j = 0; j < cnt[to[i]]; j++)
printf ("%s\n", ans[to[i]][j]);
}
return 0;
}