Solution
一般这种有两种选择的题都可以转化成最小割来做
所以我们先把所有的代价累加，求最小损失
考虑第一二种代价，分S，T连就好了。。。
连完你会发现，第三种怎么连？？？
要求在同一块儿的损失，怎么用连边表示？？
这个时候只能Orz Zsy大佬了
不同类相邻的格子
看到相邻我们想到黑白染色
那么S只连黑色的某区的边和白色的不同区的边，T相反
相邻的黑白点互连C的边，就可以直接跑了
也就大佬想得到吧
# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
# define Copy(a, b) memcpy(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(100010), __(1e6 + 10), INF(2147483647);
IL ll Read(){
RG char c = getchar(); RG ll x = 0, z = 1;
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
int n, m, ans, id[110][110], num, val[4][110][110];
int w[__], fst[_], nxt[__], to[__], cnt, S, T, lev[_], cur[_];
queue <int> Q;
IL void Add(RG int u, RG int v, RG int f, RG int _f){
w[cnt] = f; to[cnt] = v; nxt[cnt] = fst[u]; fst[u] = cnt++;
w[cnt] = _f; to[cnt] = u; nxt[cnt] = fst[v]; fst[v] = cnt++;
}
IL int Dfs(RG int u, RG int maxf){
if(u == T) return maxf;
RG int ret = 0;
for(RG int &e = cur[u]; e != -1; e = nxt[e]){
if(lev[to[e]] != lev[u] + 1 || !w[e]) continue;
RG int f = Dfs(to[e], min(w[e], maxf - ret));
ret += f; w[e ^ 1] += f; w[e] -= f;
if(ret == maxf) break;
}
return ret;
}
IL bool Bfs(){
Fill(lev, 0); lev[S] = 1; Q.push(S);
while(!Q.empty()){
RG int u = Q.front(); Q.pop();
for(RG int e = fst[u]; e != -1; e = nxt[e]){
if(lev[to[e]] || !w[e]) continue;
lev[to[e]] = lev[u] + 1;
Q.push(to[e]);
}
}
return lev[T];
}
int main(RG int argc, RG char* argv[]){
n = Read(); m = Read(); Fill(fst, -1); T = n * m + 1;
for(RG int i = 1; i <= n; ++i)
for(RG int j = 1; j <= m; ++j)
id[i][j] = ++num;
for(RG int g = 1; g <= 3; ++g)
for(RG int i = 1; i <= n; ++i)
for(RG int j = 1; j <= m; ++j){
val[g][i][j] = Read();
if(g != 3) ans += val[g][i][j];
else{
if(i > 1) ans += val[g][i][j];
if(j > 1) ans += val[g][i][j];
if(i < n) ans += val[g][i][j];
if(j < m) ans += val[g][i][j];
}
}
for(RG int i = 1; i <= n; ++i)
for(RG int j = 1; j <= m; ++j)
if((j + i) & 1) Add(S, id[i][j], val[1][i][j], 0), Add(id[i][j], T, val[2][i][j], 0);
else{
Add(S, id[i][j], val[2][i][j], 0);
Add(id[i][j], T, val[1][i][j], 0);
if(i > 1) Add(id[i][j], id[i - 1][j], val[3][i][j] + val[3][i - 1][j], val[3][i][j] + val[3][i - 1][j]);
if(j > 1) Add(id[i][j], id[i][j - 1], val[3][i][j] + val[3][i][j - 1], val[3][i][j] + val[3][i][j - 1]);
if(i < n) Add(id[i][j], id[i + 1][j], val[3][i][j] + val[3][i + 1][j], val[3][i][j] + val[3][i + 1][j]);
if(j < m) Add(id[i][j], id[i][j + 1], val[3][i][j] + val[3][i][j + 1], val[3][i][j] + val[3][i][j + 1]);
}
while(Bfs()) Copy(cur, fst), ans -= Dfs(S, INF);
printf("%d\n", ans);
return 0;
}