UVA 12506 Shortest Names (E)

Shortest Names 

In a strange village, people have very long names. For example: aaaaa, bbb and abababab.

You see, it's very inconvenient to call a person, so people invented a good way: just call a prefix of the names. For example, if you want to call `aaaaa', you can call `aaa', because no other names start with `aaa'. However, you can't call `a', because two people's names start with `a'. The people in the village are smart enough to always call the shortest possible prefix. It is guaranteed that no name is a prefix of another name (as a result, no two names can be equal).

If someone in the village wants to call every person (including himself/herself) in the village exactly once, how many characters will he/she use?

Input 

The first line contains T (T10), the number of test cases. Each test case begins with a line of one integer n ( 1n1000), the number of people in the village. Each of the following n lines contains a string consisting of lowercase letters, representing the name of a person. The sum of lengths of all the names in a test case does not exceed 1,000,000.

Output 

For each test case, print the total number of characters needed.

Sample Input 

1
3
aaaaa
bbb
abababab

Sample Output 

5

思路：先排个序，那么前缀要不一样，要最短，肯定是跟前一个比较或者后一个，取大的那个即可。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
using namespace std;
#define maxn 1010
string s[maxn];
int T, n, len1, len2, cnt;
int last[maxn], next[maxn];
int main(){
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        for(int i = 1; i <= n; i++) cin>>s[i];
        sort(s+1,s+1+n);
        cnt = 0;
        for(int i = 1; i <= n; i++){
            if(i == 1) last[i] = 0; //上一个 
            else last[i] = next[i-1];
            
            if(i == n) next[i] = 0;
            else{
                len1 = s[i].size(); len2 = s[i+1].size();
                int j;
                for( j = 0; j < len1 && j < len2; j++){
                    if(s[i][j] != s[i+1][j]){
                        break;
                    }
                }
                next[i] = j+1; //next表示i和i+1，last表示i和i-1即 next[i-1]; 
            }
            cnt += next[i]>last[i]?next[i]:last[i]; 
            
        }
        printf("%d
", cnt);
    }
    
    return 0;
}