int main(){
int n, w;
while(scanf("%d", &n) == 1) {
solver.init(n);
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++) {
scanf("%d", &w);
solver.AddEdge(i, j, w);
}
solver.solve();
int sum = 0;
for(int i = 0; i < n-1; i++) { printf("%d ", solver.Lx[i]); sum += solver.Lx[i]; } printf("%d\n", solver.Lx[n-1]);
for(int i = 0; i < n-1; i++) { printf("%d ", solver.Ly[i]); sum += solver.Ly[i]; } printf("%d\n", solver.Ly[n-1]);
printf("%d\n", sum + solver.Lx[n-1] + solver.Ly[n-1]);
}
return 0;
}
int main(){
int kase = 0;
while(scanf("%d%d", &m, &n) == 2 && m && n) {
solver.init(m*n);
int size[maxr];
for(int region = 0; region < m; region++) scanf("%d", &size[region]);
for(int program = 0; program < n; program++) {
int s[10], t[10], k;
scanf("%d", &k);
for(int i = 0; i < k; i++) scanf("%d%d", &s[i], &t[i]);
for(int region = 0; region < m; region++) {
// 计算程序p在内存区域r中的运行时间
int& time = runtime[program][region];
time = INF;
if(size[region] < s[0]) continue;
for(int i = 0; i < k; i++)
if(i == k-1 || size[region] < s[i+1]) {
time = t[i];
break;
}
// 连边X(program) -> Y(region,pos)
for(int pos = 0; pos < n; pos++)
solver.AddEdge(program, region * n + pos, -(pos + 1) * time); // 本题要求最小值，权值要取相反数
}
}
// 补完其他边
for(int i = n; i < n*m; i++)
for(int j = 0; j < n*m; j++)
solver.AddEdge(i, j, 1);
solver.solve();
printf("Case %d\n", ++kase);
print_solution();
}
return 0;
}
int main()
{
freopen("a", "r", stdin);
while (scanf("%d%d%d", &N, &M, &K) != EOF)
{
if (N == 0 && M == 0 && K == 0) break;
for (int i = 0; i < N; i++)
for (int j = 0; j < K; j++)
{
scanf("%d", &need[i][j]);
//printf("%dneed\n", need[i][j]);
}
for (int i = 0; i < M; i++)
for (int j = 0; j < K; j++)
{
scanf("%d", &have[i][j]);
//printf("%dhave\n", have[i][j]);
}
for (int k = 0; k < K; k++)
for (int i = 0; i < N; i++)
for (int j = 0; j < M; j++)
{
scanf("%d", &cost[k][i][j]);
//printf("%dcost\n", cost[k][i][j]);
}
bool ok = true;
for (int i = 0; i < K; i++)
{
int allNeed = 0, allHave = 0;
for (int j = 0; j < N; j++)
allNeed += need[j][i];
for (int j = 0; j < M; j++)
allHave += have[j][i];
if (allNeed > allHave)
{
ok = false;
printf("-1\n");
break;
}
}
if (!ok) continue;
int ans = 0;
for (int k = 0; k < K; k++)
{
int X = 0, Y = 0;
for (int i = 0; i < N; i++)
for (int j = 1; j <= need[i][k]; j++)
A[X++] = i;
for (int i = 0; i < M; i++)
for (int j = 1; j <= have[i][k]; j++)
B[Y++] = i;
km.init(X, Y);
for (int i = 0; i < X; i++)
for (int j = 0; j < Y; j++)
{
km.addEdge(i, j, -cost[k][A[i]][B[j]]);
//cout << i << " " << j << cost[k][A[i]][B[j]] << endl;
}
ans += -km.solve();
}
printf("%d\n", ans);
}
return 0;
}
