Recently, researchers on Mars have discovered N powerful atoms. All of them are different. These atoms have some properties. When two of these atoms collide, one of them disappears and a lot of power is produced. Researchers know the way every two atoms perform when collided and the power every two atoms can produce. You are to write a program to make it most powerful, which means that the sum of power produced during all the collides is maximal.
Input
There are multiple cases. The first line of each case has an integer N ( <= N <= ), which means there are N atoms: A1, A2, ... , AN. Then N lines follow. There are N integers in each line. The j-th integer on the i-th line is the power produced when Ai and Aj collide with Aj gone. All integers are positive and not larger than . The last case is followed by a in one line. There will be no more than cases including no more than large cases that N is .
Output
Output the maximal power these N atoms can produce in a line for each case.
Sample Input
Sample Output
枚举所有的状态，0表示存在，1表示不存在（已经与别的气球产生能量并且消失）。然后就是枚举所有的状态（第一个for循环），再在里面两重循环i和j，i表示不消失的那个，j表示消失的那个气球，得到了一个新的状态newS=S+（1<<j），最后就是dp[newS]=max(dp[newS],dp[S]+mp[i][j]); 边界条件是dp[S]=0, 注意这些条件（1、if(S&(1<<i)) continue; 2、if(i==j)continue; 3、if(S&(1<<j)) continue;）
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<vector>
#include<stdlib.h>
using namespace std;
#define max(a,b) (a) > (b) ? (a) : (b)
#define min(a,b) (a) < (b) ? (a) : (b)
#define ll long long
#define eps 1e-10
#define MOD 1000000007
#define N 16
#define M 1<<N
#define inf 1e12
int n;
int mp[N][N];
int dp[M];
int main()
{
while(scanf("%d",&n)==){
if(n==) break;
for(int i=;i<n;i++){
for(int j=;j<n;j++){
scanf("%d",&mp[i][j]);
}
} memset(dp,,sizeof(dp));
for(int S=;S<(<<n);S++){
for(int i=;i<n;i++){
if(S&(<<i)) continue;
for(int j=;j<n;j++){
if(i==j) continue;
if(S&(<<j)) continue;
int newS=S+(<<j);
dp[newS]=max(dp[newS],dp[S]+mp[i][j]);
}
}
}
int ans=;
for(int i=;i<(<<n);i++){
ans=max(ans,dp[i]);
}
printf("%d\n",ans);
}
return ;
}