考虑费用流，对于每一行建两个点 \(i_0, i_1\)，分别代表这一行的所有 \(0, 1\)。同样每一列建两个点 \(j_0, j_1\)。源点分别向 \(i_0, i_1\) 连流量为这一行要求的 \(0\) 或 \(1\) 的个数，费用为 \(0\)。同理连汇点。
对于一对 \((i, j)\)，我们有两种选择，让一个流量从 \(i_0 \to j_0\) 或从 \(i_1 \to j_1\)。前者在 \(a_{i, j} = 1\) 时产生 \(1\) 的费用，后者在 \(a_{i, j} = 0\) 时产生 \(1\) 的费用。
跑最小费用最大流，若不满流就无解，否则答案就是最小费用。
code
// Problem: E. Matrix Problem
// Contest: Codeforces - Educational Codeforces Round 160 (Rated for Div. 2)
// URL: https://codeforces.com/contest/1913/problem/E
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mkp make_pair
#define mems(a, x) memset((a), (x), sizeof(a))
using namespace std;
typedef long long ll;
typedef double db;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<int, int> pii;
const int maxn = 55;
const int maxm = 1000000;
const int inf = 0x3f3f3f3f;
int n, m, a[maxn][maxn], b[maxn], c[maxn];
int head[maxm], len = 1, S, T, id1[maxn][2], id2[maxn][2], nt;
struct edge {
int to, next, cap, flow, cost;
} edges[maxm];
inline void add_edge(int u, int v, int c, int f, int co) {
edges[++len].to = v;
edges[len].next = head[u];
edges[len].cap = c;
edges[len].flow = f;
edges[len].cost = co;
head[u] = len;
}
struct MCMF {
int d[maxm], cur[maxm];
bool vis[maxm];
inline void add(int u, int v, int c, int co) {
add_edge(u, v, c, 0, co);
add_edge(v, u, 0, 0, -co);
}
inline bool spfa() {
for (int i = 1; i <= nt; ++i) {
d[i] = inf;
vis[i] = 0;
}
queue<int> q;
q.push(S);
d[S] = 0;
vis[S] = 1;
while (q.size()) {
int u = q.front();
q.pop();
vis[u] = 0;
for (int i = head[u]; i; i = edges[i].next) {
edge e = edges[i];
if (e.cap > e.flow && d[e.to] > d[u] + e.cost) {
d[e.to] = d[u] + e.cost;
if (!vis[e.to]) {
vis[e.to] = 1;
q.push(e.to);
}
}
}
}
return d[T] < inf;
}
int dfs(int u, int a, int &cost) {
if (u == T || !a) {
return a;
}
vis[u] = 1;
int flow = 0, f;
for (int &i = cur[u]; i; i = edges[i].next) {
edge &e = edges[i];
if (!vis[e.to] && e.cap > e.flow && d[e.to] == d[u] + e.cost) {
if ((f = dfs(e.to, min(a, e.cap - e.flow), cost)) > 0) {
cost += f * e.cost;
e.flow += f;
edges[i ^ 1].flow -= f;
flow += f;
a -= f;
if (!a) {
break;
}
}
}
}
vis[u] = 0;
return flow;
}
inline pii solve() {
int flow = 0, cost = 0;
while (spfa()) {
for (int i = 1; i <= nt; ++i) {
cur[i] = head[i];
}
flow += dfs(S, inf, cost);
}
return mkp(flow, cost);
}
} solver;
void solve() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
scanf("%d", &a[i][j]);
}
}
S = ++nt;
T = ++nt;
for (int i = 1; i <= n; ++i) {
scanf("%d", &b[i]);
id1[i][0] = ++nt;
id1[i][1] = ++nt;
solver.add(S, id1[i][0], m - b[i], 0);
solver.add(S, id1[i][1], b[i], 0);
}
for (int i = 1; i <= m; ++i) {
scanf("%d", &c[i]);
id2[i][0] = ++nt;
id2[i][1] = ++nt;
solver.add(id2[i][0], T, n - c[i], 0);
solver.add(id2[i][1], T, c[i], 0);
}
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
solver.add(id1[i][0], id2[j][0], 1, a[i][j]);
solver.add(id1[i][1], id2[j][1], 1, a[i][j] ^ 1);
}
}
pii ans = solver.solve();
printf("%d\n", ans.fst == n * m ? ans.scd : -1);
}
int main() {
int T = 1;
// scanf("%d", &T);
while (T--) {
solve();
}
return 0;
}