Codeforces1917E - Construct Matrix

合集 - 题解(23)

1.CW初中-C0102B（加强版）（CF1720D2-Trie树）2023-12-072.CW初中-C0106D2023-12-073.Luogu-P4654-[CEOI2017] Mousetrap2023-12-074.2023.08.07模拟赛题解2023-12-085.2023.08.21 模拟赛B题2023-12-086.CW高中-C0296D2023-12-087.Codeforces Edu 154 (A-E)2023-12-088.Codeforces Pinely Round 2 (D~G)2023-12-089.Atcoder-ABC317G-Rearranging2023-12-0810.Atcoder-ARC165F-Make Adjacent2023-12-0811.Codeforces1870F-Lazy Numbers2023-12-0812.CSP2023 题解2023-12-0813.CW高中-C0389B2023-12-0814.[PA2021] Wystawa2023-12-2215.CW高中-C0443D2023-12-24

16.Codeforces1917E - Construct Matrix2023-12-25

17.Codeforces1917F - Construct Tree2023-12-2518.Atcoder Xmas Contest 2022 H - Happy Game2024-01-0719.CW高中-C0452B2024-01-0820.CW高中-C0486A2024-02-2021.QOJ #1280.Fibonacci Partition/Fibonacci性质大杂烩2024-04-0522.图论杂题2024-04-1123.QOJ #1285.Stirling Number2024-06-05

收起

Codeforces1917E - Construct Matrix

首先考虑因为 \(n\) 为偶数，所以 \(k\) 为奇数时不可能满足条件。

其次，如果 \(4|k\)，那么实际上在矩阵中一直放 \(2\times 2\) 的全为 \(1\) 的矩阵就可以了。

随后，如果 \(k \equiv 2 \mod 4\)，那么可以证明如果 \(k \ne 2 \land k \ne n^2-2\) ，那么必然有解。

我们在右下角预留一个 \(4\times 4\) 的矩阵填，先填其他位置。

然后会发现这个 \(4\times 4\) 的矩阵中，除了 \(k' = 2\) 和 \(k' = 14\) 的情况没法填，其他都有解。

又因为 \(k \ne 2 \land k \ne n^2-2\)，所以对于 \(k'=2\) 在外面少填某个 \(2\times 2\) 的矩阵即可；对于 \(k'=14\) 在外面多填某个 \(2\times 2\) 的矩阵即可。

另外 \(k=2\) 或 \(k=n^2-2\) 的情况，有解当且仅当 \(n=2\)。

代码写得很丑因为是边想边写的。

#include <bits/stdc++.h>
using namespace std;
int n, k;
int mp[1005][1005], ans[20][5][5];
int tgr[5], tgc[5];
int main (){
    int T;
    scanf ("%d", &T);
    for (int i = 0;i < (1 << 16);++ i){
        int cnt = 0;
        for (int j = 0;j < 4;++ j)
            tgr[j] = tgc[j] = 0;
        for (int j = 0;j < 16;++ j)
            if (i >> j & 1){
                ++ cnt;
                tgr[j / 4] ^= 1;
                tgc[j % 4] ^= 1;
            }
        bool tag = true;
        for (int j = 0;j < 4;++ j)
            tag = tag && tgr[j] == 0 && tgc[j] == 0;
        if (tag){
            for (int x = 0;x < 4;++ x)
                for (int y = 0;y < 4;++ y)
                    ans[cnt][x][y] = i >> ((x << 2) + y) & 1;
        }
    }
    while (T --){
        scanf ("%d%d", &n, &k);
        for (int i = 1;i <= n;++ i)
            for (int j = 1;j <= n;++ j)
                mp[i][j] = 0;
        if (k == 2 || k == n * n - 2){
            if (n == 2)
                puts ("Yes\n1 0\n0 1");
            else
                puts ("No");
        } else {
            if (k & 1)
                puts ("No");
            else {
                if (k % 4 == 0){
                    puts ("Yes");
                    for (int i = 1;i <= n && k > 0;i += 2)
                        for (int j = 1;j <= n && k > 0;j += 2){
                            k -= 4;
                            mp[i][j] = mp[i + 1][j] = mp[i][j + 1] = mp[i + 1][j + 1] = 1;
                        }
                    for (int i = 1;i <= n;++ i, puts (""))
                        for (int j = 1;j <= n;++ j)
                            printf ("%d ", mp[i][j]);
                } else {
                    if (k % 16 != 2 && k % 16 != 14){
                        int t = (k - 1) % 16 + 1;
                        k -= t;
                        for (int i = 1;i <= n && k > 0;i += 2)
                            for (int j = 1;j <= n && k > 0;j += 2)
                                if (i <= n - 4 || j <= n - 4){
                                    k -= 4;
                                    mp[i][j] = mp[i + 1][j] = mp[i][j + 1] = mp[i + 1][j + 1] = 1;
                                }
                        for (int i = 0;i < 4;++ i)
                            for (int j = 0;j < 4;++ j)
                                mp[i + n - 3][j + n - 3] = ans[t][i][j];
                        puts ("Yes");
                        for (int i = 1;i <= n;++ i, puts (""))
                            for (int j = 1;j <= n;++ j)
                                printf ("%d ", mp[i][j]);
                    } else 
                    if (k % 16 == 2){
                        int t = 6;
                        k -= t;
                        for (int i = 1;i <= n && k > 0;i += 2)
                            for (int j = 1;j <= n && k > 0;j += 2)
                                if (i <= n - 4 || j <= n - 4){
                                    k -= 4;
                                    mp[i][j] = mp[i + 1][j] = mp[i][j + 1] = mp[i + 1][j + 1] = 1;
                                }
                        for (int i = 0;i < 4;++ i)
                            for (int j = 0;j < 4;++ j)
                                mp[i + n - 3][j + n - 3] = ans[t][i][j];
                        puts ("Yes");
                        for (int i = 1;i <= n;++ i, puts (""))
                            for (int j = 1;j <= n;++ j)
                                printf ("%d ", mp[i][j]);
                    } else
                    if (k % 16 == 14){
                        int t = 10;
                        k -= t;
                        for (int i = 1;i <= n && k > 0;i += 2)
                            for (int j = 1;j <= n && k > 0;j += 2)
                                if (i <= n - 4 || j <= n - 4){
                                    k -= 4;
                                    mp[i][j] = mp[i + 1][j] = mp[i][j + 1] = mp[i + 1][j + 1] = 1;
                                }
                        for (int i = 0;i < 4;++ i)
                            for (int j = 0;j < 4;++ j)
                                mp[i + n - 3][j + n - 3] = ans[t][i][j];
                        puts ("Yes");
                        for (int i = 1;i <= n;++ i, puts (""))
                            for (int j = 1;j <= n;++ j)
                                printf ("%d ", mp[i][j]);
                    }
                }
            }
        }
    }
    return 0;
}