首先黑白染色 因为相邻的节点颜色必不同 相同颜色的节点之间没有关系(没有边)
然后Add(S,黑色点,A[i][j]) (黑色点,T,B[i][j])(S,白色点,B[i][j])(白色点,T,A[i][j])因为黑色点和白色点同属一个S/T才有额外贡献 所以这里A[i][j],B[i][j]要交换连
sum初始为矩阵里每个点的A,B之和 对于每个点 向它四周相邻的点连(u,v,C[i][j]+C[i'][j']) 注意这里黑色到白色 白色到黑色都要连 因为有属于S/T两种情况
然后一个点有X个相邻点则sum+=X*C[i][j] 因为上面连的C[i][j]+C[i'][j']容量的边其实有两条但是却不存在同时切两条的情况 最多只会切一条
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef int JQK;
int n, m;
namespace dinic {
const int MAXN = 10050;
const int MAXM = 100050;
const int INF = 1000000050;
int Head[MAXN], cur[MAXN], lev[MAXN], to[MAXM << 1], nxt[MAXM << 1], ed = 1;
int S, T, MAXP;
JQK f[MAXM << 1];
inline void addedge(int u, int v, JQK cap) {
to[++ed] = v;
nxt[ed] = Head[u];
Head[u] = ed;
f[ed] = cap;
to[++ed] = u;
nxt[ed] = Head[v];
Head[v] = ed;
f[ed] = 0;
return;
}
inline bool BFS() {
int u;
for (int i = 0; i <= MAXP + 1; i++) {
lev[i] = -1;
}
//memset(lev, -1, sizeof(lev));
queue<int>q;
lev[S] = 0;
q.push(S);
while (q.size()) {
u = q.front();
q.pop();
for (int i = Head[u]; i; i = nxt[i])
if (f[i] && lev[to[i]] == -1) {
lev[to[i]] = lev[u] + 1;
q.push(to[i]);
/*
if (to[i] == T)
{
return 1;
}
magic one way optimize
*/
}
}
for (int i = 0; i <= MAXP + 1; i++) {
cur[i] = Head[i];
}
//memcpy(cur, Head, sizeof Head);
return lev[T] != -1;
}
JQK DFS(int u, JQK maxf) {
if (u == T || !maxf) {
return maxf;
}
JQK cnt = 0, tem;
for (int &i = cur[u]; i; i = nxt[i])
if (f[i] && lev[to[i]] == lev[u] + 1) {
tem = DFS(to[i], min(maxf, f[i]));
maxf -= tem;
f[i] -= tem;
f[i ^ 1] += tem;
cnt += tem;
if (!maxf) {
break;
}
}
if (!cnt) {
lev[u] = -1;
}
return cnt;
}
JQK Dinic() {
JQK ans = 0;
while (BFS()) {
ans += DFS(S, INF);
}
return ans;
}
void init(int SS, int TT) {
for (int i = 0; i <= MAXP + 1; i++) {
Head[i] = 0;
}
ed = 1;
S = SS;
T = TT;
return;
}
int work() {
return Dinic();
}
}
int dir[4][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}};
int bl[105][105];
int sum = 0;
int A[105][105], B[105][105], C[105][105];
int main() {
int n, m;
scanf("%d %d", &n, &m);
dinic::MAXP = n * m + 5;
int s, t;
s = n * m + 1, t = s + 1;
dinic::init(s, t);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++) {
scanf("%d", &A[i][j]);
sum += A[i][j];
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++) {
scanf("%d", &B[i][j]);
sum += B[i][j];
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++) {
scanf("%d", &C[i][j]);
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
bl[i][j] = (i + j) % 2;
int now = (i - 1) * m + j;
if (bl[i][j]) {
dinic::addedge(s, now, A[i][j]), dinic::addedge(now, t, B[i][j]);
} else {
dinic::addedge(s, now, B[i][j]), dinic::addedge(now, t, A[i][j]);
}
for (int k = 0; k < 4; k++) {
int dx = i + dir[k][0];
int dy = j + dir[k][1];
if (dx >= 1 && dx <= n && dy >= 1 && dy <= m) {
int aim = (dx - 1) * m + dy;
sum += C[i][j];
dinic::addedge(now, aim, C[i][j] + C[dx][dy]);
}
}
}
}
printf("%d\n", sum - dinic::Dinic());
return 0;
}