Caddi Programming Contest 2021(AtCoder Beginner Contest 193) F.Zebraness

题目链接

Problem Statement

We have a grid with N horizontal rows and N vertical columns.
Let (i,j) denote the square at the i-th row from the top and j-th column from the left. A character ci,jc_{i,j}ci,j describes the color of (i,j).
B means the square is painted black; W means the square is painted white; ? means the square is not yet painted.

Takahashi will complete the black-and-white grid by painting each unpainted square black or white.
Let the zebraness of the grid be the number of pairs of a black square and a white square sharing a side.
Find the maximum possible zebraness of the grid that Takahashi can achieve.

Sample Input 1

2
BB
BW

Sample Output 1

Sample Input 2

3
BBB
BBB
W?W

Sample Output 2

Sample Input 3

5
?????
?????
?????
?????
?????

Sample Output 3

建图+网络流~
对一个大小为 n∗nn*nn∗n 的矩阵，显然最大值就是 2∗n∗(n−1)2*n*(n-1)2∗n∗(n−1)，那么这题就可以转化成一个最小割(或者最大流)，答案就是最大值减去最小割，建图如下：

每一个点和四周的点连边，流量为 111
对所有非 ? 的点，判断奇偶性连一条 inf 边到源点或者汇点

注意算法复杂度，EK 算法可能会超时，要用 Dicnic 算法，AC代码如下：

#include <bits/stdc++.h>using namespace std;
typedef long long ll;
const int MAXN = 1e5 + 5;
#define inf 0x3f3f3f3fstruct edge {int from, to;int cap;int flow;edge(int u, int v, int c, int f) : from(u), to(v), cap(c), flow(f) {}
};vector<edge> edges;
vector<int> G[MAXN];
int vis[MAXN];
int d[MAXN];
int cur[MAXN];
int m;
int s, t;void add(int from, int to, int cap) {edges.push_back(edge(from, to, cap, 0));edges.push_back(edge(to, from, 0, 0));int m = edges.size();G[from].push_back(m - 2);G[to].push_back(m - 1);
}bool bfs() {memset(vis, 0, sizeof(vis));queue<int> q;q.push(s);d[s] = 1;vis[s] = 1;while (!q.empty()) {int x = q.front();q.pop();for (int i = 0; i < G[x].size(); i++) {edge &e = edges[G[x][i]];if (!vis[e.to] && e.cap > e.flow) {vis[e.to] = 1;d[e.to] = d[x] + 1;q.push(e.to);}}}return vis[t];
}int dfs(int x, int a) {if (x == t || a == 0)return a;int flow = 0, f;for (int &i = cur[x]; i < G[x].size(); i++) {edge &e = edges[G[x][i]];if (d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0) {e.flow += f;edges[G[x][i] ^ 1].flow -= f;flow += f;a -= f;if (a == 0)break;}}return flow;
}int dicnic() {int ans = 0;while (bfs()) {memset(cur, 0, sizeof(cur));ans += dfs(s, inf);}return ans;
}int main() {int n;cin >> n;vector<string> c(n);for (int i = 0; i < n; i++) cin >> c[i];s = n * n + 1, t = n * n + 2;for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) {int x = i * n + j;if (c[i][j] == 'B') {if ((i + j) % 2) add(s, x, inf);else add(x, t, inf);} else if (c[i][j] == 'W') {if ((i + j) % 2) add(x, t, inf);else add(s, x, 10);}if (i < n - 1) {int y = (i + 1) * n + j;add(x, y, 1);add(y, x, 1);}if (j < n - 1) {int y = i * n + j + 1;add(x, y, 1);add(y, x, 1);}}}cout << 2 * n * (n - 1) - dicnic();return 0;
}