复习一下最小割。
考虑翻转横纵坐标和为奇数的颜色，那么就变成了，相邻两个格子，相同颜色产生 \(1\) 的贡献。
一眼 P4313 文理分科。每个格子被分到 \(S\) 表示染黑，分到 \(T\) 表示染白。对于每两个相邻格子，建两个虚点，分别表示它们都染黑或者都染白的情况。
最后跑一下最小割。
code
// Problem: F - Zebraness
// Contest: AtCoder - Caddi Programming Contest 2021(AtCoder Beginner Contest 193)
// URL: https://atcoder.jp/contests/abc193/tasks/abc193_f
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<ll, ll> pii;
const int maxn = 110;
const int maxm = 1000100;
const int inf = 0x3f3f3f3f;
const int dx[] = {0, 0, 1, -1}, dy[] = {1, -1, 0, 0};
int n, ntot, head[maxm], len = 1, S, T, id[maxn][maxn];
char s[maxn][maxn];
struct edge {
int to, next, cap, flow;
} edges[maxm];
inline void add_edge(int u, int v, int c, int f) {
edges[++len].to = v;
edges[len].next = head[u];
edges[len].cap = c;
edges[len].flow = f;
head[u] = len;
}
struct Dinic {
int d[maxm], cur[maxm];
bool vis[maxm];
inline void add(int u, int v, int c) {
add_edge(u, v, c, 0);
add_edge(v, u, 0, 0);
}
inline bool bfs() {
for (int i = 1; i <= ntot; ++i) {
vis[i] = 0;
d[i] = -1;
}
queue<int> q;
q.push(S);
vis[S] = 1;
d[S] = 0;
while (q.size()) {
int u = q.front();
q.pop();
for (int i = head[u]; i; i = edges[i].next) {
edge &e = edges[i];
if (!vis[e.to] && e.cap > e.flow) {
vis[e.to] = 1;
d[e.to] = d[u] + 1;
q.push(e.to);
}
}
}
return vis[T];
}
int dfs(int u, int a) {
if (u == T || !a) {
return a;
}
int flow = 0, f;
for (int &i = cur[u]; i; i = edges[i].next) {
edge &e = edges[i];
if (d[e.to] == d[u] + 1 && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0) {
e.flow += f;
edges[i ^ 1].flow -= f;
flow += f;
a -= f;
if (!a) {
break;
}
}
}
return flow;
}
inline int solve() {
int flow = 0;
while (bfs()) {
for (int i = 1; i <= ntot; ++i) {
cur[i] = head[i];
}
flow += dfs(S, inf);
}
return flow;
}
} solver;
void solve() {
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
scanf("%s", s[i] + 1);
for (int j = 1; j <= n; ++j) {
id[i][j] = ++ntot;
if (s[i][j] != '?' && ((i + j) & 1)) {
s[i][j] = (s[i][j] == 'B' ? 'W' : 'B');
}
}
}
S = ++ntot;
T = ++ntot;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
if (s[i][j] == 'B') {
solver.add(S, id[i][j], inf);
} else if (s[i][j] == 'W') {
solver.add(id[i][j], T, inf);
}
}
}
int s = 0;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
if ((i + j) & 1) {
continue;
}
for (int k = 0; k < 4; ++k) {
int nx = i + dx[k], ny = j + dy[k];
if (nx < 1 || nx > n || ny < 1 || ny > n) {
continue;
}
int u = ++ntot;
solver.add(S, u, 1);
solver.add(u, id[i][j], inf);
solver.add(u, id[nx][ny], inf);
u = ++ntot;
solver.add(u, T, 1);
solver.add(id[i][j], u, inf);
solver.add(id[nx][ny], u, inf);
s += 2;
}
}
}
printf("%d\n", s - solver.solve());
}
int main() {
int T = 1;
// scanf("%d", &T);
while (T--) {
solve();
}
return 0;
}