AtCoder Beginner Contest 193 F Zebraness

洛谷传送门

AtCoder 传送门

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;
}