2019 ICPC南昌 J. Prefix （字典树）

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本文介绍了一个关于字符串难度计算的问题，通过构建字典树优化查询过程，实现高效计算字符串及其前缀的难度值，并统计特定条件下前缀的数量。

题目：

yah has n strings <s1,⋯,sn>, and he generates a sequence P by two steps:

P=<s1,⋯,sn>
Replace each si with all preﬁxes of itself.

An example is:

the n strings are < aab,ab >
ﬁrst, P =< aab,ab >
then, P =< a,aa,aab,a,ab >

There are 26 positive integers d1,⋯,d26 that denote the diﬃculty to identify lowercase English letter a,b,⋯,z for yah.

The diﬃculty to identify a string strstr is (i=1∏∣str∣dstri) mod m

Now, yah wants to calculate: for each string si(i=1,⋯,n), the number of strings in P that is preﬁx of si and more diﬃcult than si.

Input

The ﬁrst line contains two integers n,mn,m(1≤n≤105,1≤m≤2×105).

The second line contains 26 positive d1,⋯,d26, 0^50≤di≤105.

The following n lines, each line contains a string, the ith lines contains si, the sum of length of all strings is not great than 2×105.

Output

The only line contains n integers, the ith integer denoting the number of strings in P that is preﬁx of si but more diﬃcult to identify for yah.

题意：

输入n个字符串，求大于字符串i 的值的前缀的个数（值= （字母*对应的权值）取余m），主要求前缀的个数

字典树模板变化

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
typedef long long ll;
using namespace std;
int root;
int w = 1;
int trie[400500][30];
int num[400500];
void build_trie(string s, int len){
	root = 0;
	for(int i = 0; i < len; i++){
		int k = s[i] - 'a';
		if(trie[root][k] == 0){
			trie[root][k] = ++w;
		}
		root = trie[root][k];
		num[root]++;
	}
}
int query(string su){
	int len = su.size();
	root = 0;
	for(int i = 0; i < len; i++){
		int k = su[i] - 'a';
		if(trie[root][k] == 0){
			return 0;
		}
		root = trie[root][k];
	} 
	return num[root];
}
int cou = 0;
int main(){
	string s[150000];
	memset(trie, 0, sizeof(trie));
	memset(num, 0, sizeof(num));
	int n, m;
	cin >> n >> m;
	int a[30];
	memset(a, 0, sizeof(a));
	for(int i = 0; i < 26; i++){
		cin >> a[i];
	}
	for(int i = 0; i < n; i++){
		cin >> s[i];
		int x = s[i].size();
		build_trie(s[i], x);
		
	}
	
   for(int i = 0; i < n; i++){
   	int len = s[i].size();
   	ll tmp = 1;
		ll  cnt = 0;
		ll k = 1;
		string su;
		for(int j = 0; j < len; j++){
			k = (k * a[s[i][j]-'a']) % m;  // 字符串 i的值
		} 
   	ll ans = 0;
   	for(int j = 0; j < len; j++){
   		su = s[i].substr(0, j+1);
   		tmp = (tmp * a[s[i][j]- 'a'] ) % m;  字符串i的前缀的值
   		if(tmp > k){
   			ans = ans + query(su);   // 查找个数
			}
		}
		cout << ans << " ";
	}
	cout << endl;
	return 0;
}

用暴力的方法，用map存字符前缀出现的次数

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <map>
typedef long long ll;
using namespace std;
map<string, ll> ma;
ll a[30], k[200005];
int main(){
   int n, m;
   cin >> n >> m;
   string s[100005];
	memset(a, 0, sizeof(a));
	for(int i = 0; i < 26; i++){
		cin >> a[i];
	}
	for(int i = 1; i <= n; i++){
		k[i] = 1;
		cin >> s[i];
		int len = s[i].size();
		for(int j = 0; j < len; j++){
			string ss;
			ss = s[i].substr(0,j+1);
			ma[ss]++;
			k[i] = (k[i] * a[s[i][j]-'a']) % m;
		}
	}
	for(int i = 1; i <= n; i++){
		int len = s[i].size();
		ll w = 1, ans = 0;
		for(int j = 0; j < len; j++){
			w = (w * a[s[i][j]-'a']) % m;
			if(w > k[i]){
				string ss;
				ss = s[i].substr(0,j+1);
				ans = ans + ma[ss];
			}
		}
		cout << ans << " ";
	}
	cin >> n;
   return 0;
}