题目描述
数组内容串了~
样例
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
//#define int long long
const int N = 20 , M = 1 << N;
int n;
int w[N][N];
int f[M][N];
/*#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 20, M = 1 << N;
int n;
int w[N][N];
int f[M][N];*/
int main()
{
/*cin >> n;
for(int i = 0 ; i < n ; i++)
for(int j = 0 ; j < n ; j++)
cin >> w[i][j];
memset(f,0x3f, sizeof f);
f[1][0] = 0;*/
cin >> n;
for (int i = 0; i < n; i ++ )
for (int j = 0; j < n; j ++ )
cin >> w[i][j];
memset(f, 0x3f, sizeof f);
f[1][0] = 0;
for (int i = 0; i < 1 << n; i ++ )
for (int j = 0; j < n; j ++ )
if (i >> j & 1)
for (int k = 0; k < n; k ++ )
if (i >> k & 1)
f[i][j] = min(f[i][j], f[i - (1 << j)][k] + w[k][j]);
/*for(int i = 0 ; i < 1 << n ; i++)
for(int j = 0 ; j < n ; j++)
if(i >> j & 1)
for(int k = 0 ; k < n ; k ++)
if(i >> k & 1)
f[i][j] = min(f[i][j],f[i-(1<<j)][k] + w[k][j]);*/
cout << f[(1 << n) - 1][n - 1];
return 0;
/*cout << f[(1 << n) - 1][n-1];//遍历所有情况
return 0;*/
}