BZOJ_3448_[Usaco2014 Feb]Auto-complete_Trie树

BZOJ_3448_[Usaco2014 Feb]Auto-complete_Trie

Description

Bessie the cow has a new cell phone and enjoys sending text messages, although she keeps making spelling errors since she has trouble typing on such a small screen with her large hooves. Farmer John has agreed to help her by writing an auto-completion app that takes a partial word and suggests how to complete it. The auto-completion app has access to a dictionary of W words, each consisting of lowercase letters in the range a..z, where the total number of letters among all words is at most 1,000,000. The app is given as input a list of N partial words (1 <= N <= 1000), each containing at most 1000 lowercase letters. Along with each partial word i, an integer K_i is also provided, such that the app must find the (K_i)th word in alphabetical order that has partial word i as a prefix. That is, if one ordered all of the valid completions of the ith partial word, the app should output the completion that is (K_i)th in this sequence.

XYW在和他的男人聊天的时候经常会因为打字速度太慢而被鄙视，于是他想开发一种自动补全软件。开始他有一个包含了w个词的字典（总长度不大于10^6）。接下来他想知道在字典中可以由给定的字符串s加上若干个字母得到的字符串字典序第k大的编号是多少？有n组询问（n<=10^3，s长度<=10^3)

Input

* Line 1: Two integers: W and N.

 * Lines 2..W+1: Line i+1: The ith word in the dictionary.

* Lines W+2..W+N+1: Line W+i+1: A single integer K_i followed by a partial word. 

Output

 * Lines 1..N: Line i should contain the index within the dictionary (an integer in the range 1..W) of the (K_i)th completion (in alphabetical order) of the ith partial word, or -1 if there are less than K_i completions.

Sample Input

10 3
dab
ba
ab
daa
aa
aaa
aab
abc
ac
dadba
4 a
2 da
4 da

Sample Output

3
1
-1

---

并没有想到排序加二分。。

直接对所有串建立trie树，维护子树siz，这样我们查询的时候从a开始查然后不断减K即可。

BZ上面加强了数据，直接这么做会T死。

考虑优化：对于Trie树上那些非叶子节点且只有一个儿子的点，可以直接合并到下面的结点上去。。

这个玩意是我受到后缀树的启发bb出来的，复杂度也不会证，应该可以卡掉。

BZ的数据字符总数可能超过1000000。开2100000就够了(我这个做法是这样

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define N 2100050
int t[N],ch[N][26],W,n,cnt=1,flg[N],fa[N],is[N],vis[N];
char w[N];
void insert(int idx) {
    int p=1,i;  
    for(i=1;w[i];i++) {
        int &k=ch[p][w[i]-'a'];
        if(!k) k=++cnt;
        p=k;
        t[p]++; 
        if(!flg[p]&&!vis[p]) flg[p]=idx;
        else if(!is[p]) flg[p]=0,vis[p]=1;
    }
    is[p]=1;
    flg[p]=idx;
}
int query(int p,int k) {
    if(k==1&&flg[p]) return flg[p];
    int i;
    int now=is[p];
    for(i=0;i<26;i++) {
        now+=t[ch[p][i]];
        if(now>=k) {
            return query(ch[p][i],k-now+t[ch[p][i]]);
        }
    }
    return flg[p];
}
int main() {
    scanf("%d%d",&W,&n);
    int i,j,sum=0;
    for(i=1;i<=W;i++) {
        scanf("%s",w+1);
        insert(i);
        // sum+=strlen(w+1);
    }
    // printf("%d
",sum); return 0;
    int K;
    for(i=1;i<=n;i++) {
        scanf("%d%s",&K,w+1);
        int p=1,flg=1;
        for(j=1;w[j];j++) {
            if(!ch[p][w[j]-'a']) {flg=0; break;}
            p=ch[p][w[j]-'a'];
        }
        if(!flg||t[p]<K) puts("-1");
        else printf("%d
",query(p,K)); 
    }
}