AC Diary-Solitaire game Codevs 1051

1051 Solitaire Games

time limit: 1 sspace limit: 128000 KBtitle level: Diamonds DiamondTitle Description Description

Given the n words, the order has been sorted by length. If a word i is a prefix for a word j, I->j is counted as a solitaire (two identical words cannot be counted as solitaire).

Your task is: To find the longest dragon for the words entered.

Enter a description Input Description

The first behavior N (1<=n<=105). The following n lines, one word per line (made up of lowercase), have been sorted by length. (<50 per word length)

Output description Output Description

Only one number, for the length of the longest dragon.

Sample input Sample Input

5

I

A

Int

Able

Inter

Sample output Sample Output

3

Data range and Tips Data Size & Hint

1<=n<=105

Ideas:

Use a hash table to save all the words.

and compare it with a hash table.

Then the linear structure compares whether the complexity of the prefix is n^2

However n<=100000

Time-outs steady.

So, how to optimize it?

Consider transforming the linear comparison into a tree

Establish a root node

We're looking for the word prefix from the root node.

If there is no prefix for this word

The word is grown on the root node as a leaf node.

If you find the

Then jump into this leaf node

Continue to look for prefixes from this leaf node

Until it is not found, and then the word as a new leaf node grows on the current node

This will optimize a lot of

Easy AC

Come on, on the code:

#include <cstdio>#include<string>#include<cstring>#include<iostream>#defineMoD 10000007using namespacestd;structNode {int  from, To,dis,next;};structNode edge[100001];intn,len[100001],hash[100001][Wuyi],dp[100001];inthead[100001],num,ans=0;stringword[100001];inlinevoidEdge_add (int  from,intTo ) {num++; Edge[num].to=to ; Edge[num]. from= from; Edge[num].next=head[ from]; head[ from]=num;}voidDfsintNowintPosintBe ) {ans=Max (Now,ans);  for(intI=head[be];i;i=Edge[i].next) {        if(len[edge[i].to]<Len[pos]) {            if(hash[edge[i].to][len[edge[i].to]-1]==hash[pos][len[edge[i].to]-1]) {DFS ( now+1, pos,edge[i].to); return ; }}} edge_add (Be,pos);}intMain () {len[0]=1; scanf ("%d",&N);  for(intI=1; i<=n;i++) {Dp[i]=1; CIN>>Word[i]; Len[i]=word[i].length ();  for(intj=0; j<len[i];j++) {Hash[i][j]=word[i][j]-'a'+1; if(j) hash[i][j]+= (hash[i][j-1]* in)%MoD; HASH[I][J]%=MoD; } DFS (1I0); } cout<<ans<<Endl; return 0;}

AC Diary-Solitaire game Codevs 1051