poj 2334 Simple prefix compression(胡搞)

Simple prefix compression

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 1861 Accepted: 813

Description

Many databases store the data in the character fields (and especially indices) using prefix compression. This technique compresses a sequence of strings A1, ..., A_N by the following method: if there are strings Ai = a_i,1a_i,2...a_i,p and A_{i + 1} = a_i+1,1a_i+1,2...a_i+1,q 
such that for some j ≤ min(p, q) a_i,1 = a_i+1,1, a_i,2 = a_i+1,2, ... a_i,j = a_i+1,j, then the second string is stored as [j]a_i+1,j+1a_i+1,j+2... a_i+1,q, where [j] is a single character with code j. 

If j = 0, that is, strings do not have any common prefix, then the second string is prefixed with zero byte, and so the total length actually increases. 


Constraints 
1 ≤ N ≤ 10000, 1 ≤ length(Ai) ≤ 255.

Input

First line of input contains integer number N, with following N lines containing strings A1 ... A_N

Output

Output must contain a single integer -- minimal total length of compressed strings.

Sample Input

3abcatestatext

Sample Output

11

tips：马德，至今未看懂题意；看着别人的题解慢慢的胡搞；

具体来说每一次都统计与前面一个字符串的前缀数量；然后将长度-前缀数量的长度累加起来；然后胡搞没看懂题

#include<iostream>#include<cstring>#include<string>using namespace std;int n;string last,now; int main(){cin>>n;int ans=0;while(n--){last=now;cin>>now;int sz=0;for(int i=0;i<last.size()&&i<now.size();i++){if(now[i]!=last[i])break;sz++;//统计前缀长度 }ans+=(now.size()-sz+1);//非前缀长度； }cout<<ans-1<<endl; return 0; }