Caddi Programming Contest 2021(AtCoder Beginner Contest 193) F.Zebraness_atcoder beginner contest 194

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 $c_{i,j}$ 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
1
2
3

Sample Output 1

2
1

Sample Input 2

3
BBB
BBB
W?W
1
2
3
4

Sample Output 2

4
1

Sample Input 3

5
?????
?????
?????
?????
?????
1
2
3
4
5
6

Sample Output 3

40
1

建图+网络流~
对一个大小为 $n\ast n$ 的矩阵，显然最大值就是 $2\ast n\ast (n-1)$，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 $1$
• 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 0x3f3f3f3f

struct 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;
}
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