2391: Words n个字符串最多能连多长

2391: Words

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	532	95	Standard

Io and Ao are playing a word game. They say a word in the dictionary by turns. The word in the dictionary only contains lowercase letters. And the end character of the former said word should be the same as the start character of the current said word. They can start the game from any word in the dictionary. Any word shouldn't be said twice. Now, we define the complexity of the game that is the sum length of all words said in the game. Give you a dictionary, can you tell me the max complexity of this word game?

Input

The first line contains a single positive integer n(0 < n <=12). Next n lines are n words in the dictionary. The length of each word will not exceed 100.

Output

A single integer represents the complexity of the game.

Sample Input

3
joj
jlu
acm
6
cctv
redcode
lindong
we
love
programming
3
daoyuanlee
come
on

Sample Output

6
11
10

Problem Source: provided by loon

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
using namespace std;
int n,cnt,vis[20];
char a[20][150];
int ok;
void dfs(int id,int now,int num)
{
if(now>cnt) cnt=now;
if(num==n) {ok=1;return ;}
for(int i=0;i<n;i++)
{
if(ok) return ;
if(vis[i]==0&&(num==0||a[id][strlen(a[id])-1]==a[i][0]))
{
vis[i]=1;
dfs(i,now+strlen(a[i]),num+1);
vis[i]=0;
}
}
}
int main()
{
while(scanf("%d",&n)==1)
{
cnt=0;ok=0;
for(int i=0;i<n;i++) scanf("%s",a[i]);
memset(vis,0,sizeof(vis));
dfs(0,0,0);
printf("%d/n",cnt);
}
return 0;
}
/*
4
abbc
abc
ba
cg
8
*/