ABC347G 题题解

还算是比较经典了。

首先我们注意到一个性质：\(1 + 3 + \cdots + n = n ^ 2\)。所以我们可以把平方拆开。

然后容易证明 \(a_{i, j}\) 填 \(1\) 一定比填 \(0\) 不劣。

我们可以把 \(a_{i, j}\) 拆成 \(4\) 个点，然后我们想到了最小割。

构造网络：

• \(a_{i, j, x} \leftarrow a_{i, j, x + 1}\)，权重为 \(+\infty\)。

• 若 \(a_{i, j} != 0\)，\(\forall x \neq a_{i, j} - 1\)，\(a_{i, j, x} \rightarrow a_{i, j, x + 1}\)，权重为 \(+\infty\)。

• 对于相邻的两个点 \(A, B\)，\(\forall x > y, a_{A, x} \rightarrow a_{B, y}\)， 权重为 \(2\)。

• 对于相邻的两个点 \(A, B\)，\(\forall x, a_{A, x} \rightarrow a_{B, x}\)， 权重为 \(1\)。

容易证明这个网络的最小割即为所求。

按照 Dinic 跑最大流，残量网络即为最小割。时间复杂度 \(\mathcal{O}(F\times m) = \mathcal{O}(n^4 \times d^4)\)，能过。

/*******************************
| Author:  DE_aemmprty
| Problem: G - Grid Coloring 2
| Contest: AtCoder - AtCoder Beginner Contest 347
| URL:     https://atcoder.jp/contests/abc347/tasks/abc347_g
| When:    2024-04-02 12:29:05
| 
| Memory:  1024 MB
| Time:    2000 ms
*******************************/
#include <bits/stdc++.h>
using namespace std;

long long read() {
    char c = getchar();
    long long x = 0, p = 1;
    while ((c < '0' || c > '9') && c != '-') c = getchar();
    if (c == '-') p = -1, c = getchar();
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return x * p;
}

const int N = 1607;

struct Edge {
    int v;
    long long w;
    int nxt;
} e[N * N];

int n, P;
int hd[N], tot;

void add(int u, int v, long long w) {
    e[++ tot] = {v, w, hd[u]};
    hd[u] = tot;
}

namespace Dinic {
    long long ans, dis[N], now[N];
    int grp[N];
    bool Build(int x, int t) {
        queue <int> q;
        fill(dis + 1, dis + P + 1, 2e18);
        q.push(x); dis[x] = 0, now[x] = hd[x];
        while (!q.empty()) {
            int u = q.front(); q.pop();
            if (u == t) return true;
            for (int i = hd[u]; i > 1; i = e[i].nxt) {
                int v = e[i].v, w = e[i].w;
                if (w != 0 && dis[v] == 2e18) {
                    dis[v] = dis[u] + 1;
                    now[v] = hd[v];
                    q.push(v);
                }
            }
        }
        return false;
    }
    long long update(int x, int t, long long sum) {
        if (x == t) return sum;
        long long res = 0;
        for (int i = now[x]; i > 1 && sum; i = e[i].nxt) {
            now[x] = i;
            long long v = e[i].v, w = e[i].w;
            if (w > 0 && dis[v] == dis[x] + 1) {
                long long p = update(v, t, min(sum, w));
                if (!p) dis[v] = 2e18;
                e[i].w -= p;
                e[i ^ 1].w += p;
                res += p;
                sum -= p;
            }
        }
        return res;
    }
    long long Dinic(int s, int t) {
        ans = 0;
        while (Build(s, t))
            ans += Dinic::update(s, t, 2e18);
        return ans;
    }
    void MinCut(int s, int t) {
        for (int i = 1; i <= P; i ++)
            if (dis[i] != 2e18)
                grp[i] = 1;
            else
                grp[i] = 0;
    }
    void Add(int u, int v, long long w) {
        add(u, v, w);
        add(v, u, 0);
    }
}

int a[27][27];
int col[27][27][17];

void solve() {
    n = read();
    for (int i = 1; i <= n; i ++) for (int j = 1; j <= n; j ++)
        a[i][j] = read();
    tot = 1; P = 2;
    for (int i = 1; i <= n; i ++) for (int j = 1; j <= n; j ++) {
        // col[i][j][0] = 1, col[i][j][5] = 2;
        for (int x = 1; x <= 4; x ++) col[i][j][x] = ++ P;
    }
    for (int i = 1; i <= n; i ++) for (int j = 1; j <= n; j ++)
        for (int x = 2; x <= 4; x ++)
            Dinic::Add(col[i][j][x], col[i][j][x - 1], 2e18);
    for (int i = 1; i <= n; i ++) for (int j = 1; j <= n; j ++)
        if (a[i][j]) {
            if (a[i][j] > 1) Dinic::Add(1, col[i][j][a[i][j] - 1], 2e18);
            if (a[i][j] < 5) Dinic::Add(col[i][j][a[i][j]], 2, 2e18);
        }
    for (int i = 1; i <= n; i ++) for (int j = 1; j <= n; j ++)
        for (int x = 1; x <= 4; x ++) for (int y = 1; y <= x; y ++) {
            if (y < x) {
                if (i < n) Dinic::Add(col[i][j][x], col[i + 1][j][y], 2);
                if (i > 1) Dinic::Add(col[i][j][x], col[i - 1][j][y], 2);
                if (j < n) Dinic::Add(col[i][j][x], col[i][j + 1][y], 2);
                if (j > 1) Dinic::Add(col[i][j][x], col[i][j - 1][y], 2);
            } else {
                if (i < n) Dinic::Add(col[i][j][x], col[i + 1][j][x], 1);
                if (i > 1) Dinic::Add(col[i][j][x], col[i - 1][j][x], 1);
                if (j < n) Dinic::Add(col[i][j][x], col[i][j + 1][x], 1);
                if (j > 1) Dinic::Add(col[i][j][x], col[i][j - 1][x], 1);
            }
        }
    Dinic::Dinic(1, 2);
    Dinic::MinCut(1, 2);
    for (int i = 1; i <= n; i ++) {
        for (int j = 1; j <= n; j ++) {
            int px = 1;
            for (int x = 1; x <= 4; x ++)
                px += Dinic::grp[col[i][j][x]];
            cout << px << ' ';
        }
        cout << '\n';
    }
}

signed main() {
    int t = 1;
    while (t --) solve();
    return 0;
}