#include 
     
      #include 
      
       #define M 310#define inf 0x3f3f3f3fint n,nx,ny;int link[M],lx[M],ly[M],slack[M];    //lx,ly为顶标，nx,ny分别为x点集y点集的个数int visx[M],visy[M],w[M][M];int DFS(int x){    visx[x] = 1;    for (int y = 1;y <= ny;y ++)    {        if (visy[y])            continue;        int t = lx[x] + ly[y] - w[x][y];        if (t == 0)       //        {            visy[y] = 1;            if (link[y] == -1||DFS(link[y]))            {                link[y] = x;                return 1;            }        }        else if (slack[y] > t)  //不在相等子图中slack 取最小的            slack[y] = t;    }    return 0;}int KM(){    int i,j;    memset (link,-1,sizeof(link));    memset (ly,0,sizeof(ly));    for (i = 1;i <= nx;i ++)            //lx初始化为与它关联边中最大的        for (j = 1,lx[i] = -inf;j <= ny;j ++)            if (w[i][j] > lx[i])                lx[i] = w[i][j];    for (int x = 1;x <= nx;x ++)    {        for (i = 1;i <= ny;i ++)            slack[i] = inf;        while (1)        {            memset (visx,0,sizeof(visx));            memset (visy,0,sizeof(visy));            if (DFS(x))     //若成功（找到了增广轨），则该点增广完毕。进入下一个点的增广                break;  //若失败（没有找到增广轨），则须要改变一些点的标号，使得图中可行边的数量添加。                        //方法为：将全部在增广轨中（就是在增广过程中遍历到）的X方点的标号全部减去一个常数d，                        //全部在增广轨中的Y方点的标号全部加上一个常数d            int d = inf;            for (i = 1;i <= ny;i ++)                if (!visy[i]&&d > slack[i])                    d = slack[i];            for (i = 1;i <= nx;i ++)                if (visx[i])                    lx[i] -= d;            for (i = 1;i <= ny;i ++)  //改动顶标后，要把全部不在交错树中的Y顶点的slack值都减去d                if (visy[i])                    ly[i] += d;                else                    slack[i] -= d;        }    }    int res = 0;    for (i = 1;i <= ny;i ++)        if (link[i] > -1)            res += w[link[i]][i];    return res;}int main (){    int i,j;    while (scanf ("%d",&n)!=EOF)    {        nx = ny = n;      //  memset (w,0,sizeof(w));        for (i = 1;i <= n;i ++)            for (j = 1;j <= n;j ++)                scanf ("%d",&w[i][j]);        int ans = KM();        printf ("%d\n",ans);    }    return 0;}