[省选联考 2021 A 卷] 矩阵游戏

比较有意思的一道题。

首先不考虑 \(a_{i, j}\) 的限制随便求，然后开始调整。你发现对于每行，奇数列 \(+x\) 偶数列 \(-x\) 不会变，列也同理，假设第 \(i\) 行的 \(x\) 为 \(r_i\), 第 \(j\) 列的 \(x\) 为 \(c_j\)。然后考虑对 \(a\) 加上一个矩阵 \(A\)，\(A_{i, j} = (-1)^{2\mid i+1}r_i - (-1)^{2\mid j+1}c_j\)。然后你发现这样会有加和，不能跑差分约束，调整一下，\(\forall 2 \mid i, r_i \leftarrow -r_i, \forall 2 \nmid j, c_j \leftarrow -c_j\)，就都是差的形式了。然后差分约束即可，时间复杂度 \(\mathcal{O}(nm(n+m))\)。

点击查看代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
// typedef __int128 i128;
typedef pair<int, int> pii;
const int N = 3e2 + 10, V = 1e6, mod = 998244353;
template<typename T>
void dbg(const T &t) { cout << t << endl; }
template<typename Type, typename... Types>
void dbg(const Type& arg, const Types&... args) {
    cout << arg << ' ';
    dbg(args...);
}
namespace Loop1st {
int n, m, b[N][N], a[N][N];
ll dis[N << 1];
vector<pii>e[N << 1];
void main() {
    memset(dis, 0x3f, sizeof dis); memset(a, 0, sizeof a);
    cin >> n >> m;
    for (int i = 0; i <= n + m; i++) e[i].clear();
    for (int i = 1; i <= n + m; i++) e[0].push_back({i, 0});
    for (int i = 1; i < n; i++) for (int j = 1; j < m; j++) cin >> b[i][j];
    for (int i = n - 1; i; i--) {
        for (int j = m - 1; j; j--) {
            a[i][j] = b[i][j] - a[i + 1][j] - a[i][j + 1] - a[i + 1][j + 1];
        }
    }
    for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) {
        if ((i + j) & 1) e[n + j].push_back({i, a[i][j]}), e[i].push_back({n + j, V - a[i][j]});
        else e[i].push_back({n + j, a[i][j]}), e[n + j].push_back({i, V - a[i][j]});
    }
    dis[0] = 0;
    int t = 1;
    for (t = 1; t <= n + m + 2; t++) {
        bool ok = 0;
        for (int u = 0; u <= n + m; u++) {
            for (auto [v, w] : e[u]) if (dis[v] > dis[u] + w) {
                dis[v] = dis[u] + w;
                ok = 1;
            }
        }
        if (!ok) break;
    }
    if (t > n + m + 2) {
        cout << "NO\n";
        return ;
    }
    cout << "YES\n";
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            // dbg("###", i, j, dis[i], dis[n + j], a[i][j]);
            if ((i + j) & 1) cout << a[i][j] + dis[n + j] - dis[i] << " \n"[j == m];
            else cout << a[i][j] + dis[i] - dis[n + j] << " \n"[j == m];
        }
    }
}

}
int main() {
    // freopen("data.in", "r", stdin);
    // freopen("data.out", "w", stdout);
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int T = 1;
    cin >> T;
    while (T--) Loop1st::main();
    return 0;
}