【USACO14FEB】洛谷2237 Auto-complete

题目描述

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.

贝西新买了手机，打字不方便，请设计一款应用，帮助她快速发消息。

字典里有W（W<=30000）个小写字母构成的单词，所有单词的字符总数量不超过1,000,000，这些单词是无序的。现在给出N(1 <= N

<=

1000)个询问，每个询问i包含一个的字符串s_i（每个字符串最多包含1000个字符）和一个整数K_i，对于所有以s_i为前缀的单词，其中按字典序排序后的第K_i个单词，求该单词在原字典里的序号。

输入输出格式 输入格式：

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.

输出格式：

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.

把每个字符串插入trie中，顺便记录下每个结点经过的字符串个数。

对于每个询问在每个节点上按照个数找就可以了，有点类似字典序解码。

#include<cstdio>
#include<cstring>
const int a_s=26;
int trie[1000002][27],num[1000010],tot,size[1000010];
char s[1000010];
int main()
{
int i,j,k,m,n,p,q,x,y,z,l,u;
bool flag;
scanf("%d%d",&n,&m);
for (i=1;i<=n;i++)
{
scanf("%s",s+1);
l=strlen(s+1);
p=0;
for (j=1;j<=l;j++)
{
size[p]++;
x=s[j]-'a'+1;
if (!trie[p][x]) trie[p][x]=++tot;
p=trie[p][x];
}
size[p]++;
num[p]=i;
}
while (m--)
{
scanf("%d%s",&u,s+1);
l=strlen(s+1);
p=0;
flag=1;
for (i=1;i<=l;i++)
{
x=s[i]-'a'+1;
if (!trie[p][x])
{
flag=0;
break;
}
p=trie[p][x];
}
if (flag==0||size[p]<u)
{
printf("-1\n");
continue;
}
while (1)
{
if (num[p]) u--;
if (!u)
{
printf("%d\n",num[p]);
break;
}
for (i=1;i<=a_s;i++)
if (trie[p][i])
{
x=trie[p][i];
if (size[x]<u) u-=size[x];
else
{
p=x;
break;
}
}
}
}
}