[二分图&最小割]

#define MAXN 110
#include <bits/stdc++.h>
 
using namespace std;
const int inf = 0x7fffffff / 2;
const int dx[4] = {-1, 1, 0, 0}, dy[4] = {0, 0, -1, 1};
 
int n, m, S, T, a[MAXN][MAXN], b[MAXN][MAXN], c[MAXN][MAXN];
int h[MAXN*MAXN], cnt = 1;
struct Edge { int to, nxt, w; } edge[2000010];
void addedge(int u, int v, int w) {
	edge[++ cnt] = (Edge){v, h[u], w}; h[u] = cnt;
	edge[++ cnt] = (Edge){u, h[v], 0}; h[v] = cnt;
}
 
int que[MAXN*MAXN], d[MAXN*MAXN];
bool BFS() {
	int head = 0, tail = 0;
	for(int i = S ; i <= T ; ++ i) d[i] = -1;
	que[tail ++] = S, d[S] = 0;
	while(head != tail) {
		int u = que[head ++];
		for(int i = h[u] ; i ; i = edge[i].nxt) {
			if(edge[i].w == 0) continue;
			int v = edge[i].to;
			if(d[v] == -1) d[v] = d[u] + 1, que[tail ++] = v;
		}
	} return d[T] != -1;
}
 
int DFS(int x, int a) {
	if(x == T || a == 0) return a;
	int used = 0, f;
	for(int i = h[x] ; i ; i = edge[i].nxt) {
		int v = edge[i].to;
		if(d[v] == d[x] + 1) {
			f = DFS(v, min(a-used, edge[i].w));
			edge[i].w -= f;
			edge[i^1].w += f;
			used += f;
			if(used == a) return used;
		}
	}
	if(!used)d[x] = -1;
	return used;
}
 
int Dinic() {
	int ret = 0;
	while(BFS())
		ret += DFS(S, inf);
	return ret;
}
 
int main() {
	freopen("nt2011_land.in", "r", stdin);
	freopen("nt2011_land.out", "w", stdout);
	scanf("%d%d", &n, &m);
	for(int i = 1 ; i <= n ; ++ i)
		for(int j = 1 ; j <= m ; ++ j)
			scanf("%d", &a[i][j]);
	for(int i = 1 ; i <= n ; ++ i)
		for(int j = 1 ; j <= m ; ++ j)
			scanf("%d", &b[i][j]);
	for(int i = 1 ; i <= n ; ++ i)
		for(int j = 1 ; j <= m ; ++ j)
			scanf("%d", &c[i][j]);
	static int id[MAXN][MAXN], idfclock = 0;
	for(int i = 1 ; i <= n ; ++ i)
		for(int j = 1 ; j <= m ; ++ j)
			id[i][j] = ++ idfclock;
	S = 0, T = ++ idfclock;
	for(int i = 1 ; i <= n ; ++ i)
		for(int j = 1 ; j <= m ; ++ j) {
			if(i + j & 1) {
				addedge(S, id[i][j], a[i][j]);
				addedge(id[i][j], T, b[i][j]);
			}
			else {
				addedge(S, id[i][j], b[i][j]);
				addedge(id[i][j], T, a[i][j]);
			}
			for(int k = 0 ; k < 4 ; ++ k) {
				int x = i + dx[k], y = j + dy[k];
				if(id[x][y])addedge(id[i][j], id[x][y], c[i][j] + c[x][y]);				
			}
		}
	int ans = 0;
	for(int i = 1 ; i <= n ; ++ i)
		for(int j = 1 ; j <= m ; ++ j) {
			int cn = 0;
			for(int k = 0 ; k < 4 ; ++ k) {
				int x = i + dx[k], y = j + dy[k];
				if(id[x][y]) cn ++;
			}
			ans += a[i][j] + b[i][j] + c[i][j] * cn;
		}
 
	printf("%d\n", ans - Dinic());
	return 0;
}