2334:Simple prefix compression

总时间限制: 2000ms 内存限制: 65536kB
描述
Many databases store the data in the character fields (and especially indices) using prefix compression. This technique compresses a sequence of strings A1, …, AN by the following method: if there are strings Ai = ai,1ai,2…ai,p and Ai + 1 = ai+1,1ai+1,2…ai+1,q
such that for some j ≤ min(p, q) ai,1 = ai+1,1, ai,2 = ai+1,2, … ai,j = ai+1,j, then the second string is stored as [j]ai+1,j+1ai+1,j+2… ai+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.
输入
First line of input contains integer number N, with following N lines containing strings A1 … AN
输出
Output must contain a single integer – minimal total length of compressed strings.
样例输入
3
abc
atest
atext
样例输出
11

#include<iostream>#include<string.h>using namespace std;//http://bailian.openjudge.cn/practice/2334///原来这个题就是一个字符串和它相邻的下一个字符串进行比较//本来还以为是要全部比较，那就应该是trie树了吧 char a[11000][300];int n,cnt=0,len;int main(){    cin>>n;    for(int i=0;i<n;i++){        cin>>a[i];        len=strlen(a[i]);        cnt+=len;//先都加起来     }    for(int i=0;i<n-1;i++){        for(int j=0;j<min(strlen(a[i]),strlen(a[i+1]));j++){            if(a[i][j]==a[i+1][j]){                cnt--;//相同字符就减一             }            else break;        }        cnt++;//加j这个字符     }    cout<<cnt<<endl;}