Codeforces Round #454 (Div. 2, based on Technocup 2018 Elimination Round 4) D. Seating of Students [模拟]

题意：给一个n行m列的矩阵，原矩阵按数字顺序从第一行开始向后填充，寻找一种方案使得原矩阵中相邻的数字在新矩阵中都不相邻。

分析：没有什么技术含量的一道构造题，试试就可以发现方案，很多人是用随机化过的这道题，但是对速度要求较高。比较稳定的方法是分两行或三行叉开放置，并且在第二行放置的时候先放两个最大的再按顺序放，例如：

可以转化为

可以转化为

只需要用合理的排列顺序就可以处理好行列大于3的所有情况

还需要特判1行、1*1、3*3的特殊情况，其他情况都无法成立。当然还有其他很多种排列方式，为了答案稳定，写的比较繁琐分了n<m和n>m，大多数代码基本都是复制粘贴，题目也没有什么算法，中间有很多地方可以优化。

代码：

  1 #define _CRT_SECURE_NO_DEPRECATE
  2 #pragma comment(linker, "/STACK:102400000,102400000")
  3 #include<iostream>  
  4 #include<cstdio>  
  5 #include<fstream>  
  6 #include<iomanip>
  7 #include<algorithm>  
  8 #include<cmath>  
  9 #include<deque>  
 10 #include<vector>
 11 #include<bitset>
 12 #include<queue>  
 13 #include<string>  
 14 #include<cstring>  
 15 #include<map>  
 16 #include<stack>  
 17 #include<set>
 18 #include<functional>
 19 #define pii pair<int, int>
 20 #define mod 1000000007
 21 #define mp make_pair
 22 #define pi acos(-1)
 23 #define eps 0.00000001
 24 #define mst(a,i) memset(a,i,sizeof(a))
 25 #define all(n) n.begin(),n.end()
 26 #define lson(x) ((x<<1))  
 27 #define rson(x) ((x<<1)|1) 
 28 #define inf 0x3f3f3f3f
 29 typedef long long ll;
 30 typedef unsigned long long ull;
 31 using namespace std;
 32 const int maxn = 1e5 + 5;
 33 vector<int>a[maxn];
 34 vector<int>ans[maxn];
 35 int n, m, cas = 0;
 36 
 37 void change2(int t)
 38 {
 39     ans[t].resize(m+1);
 40     ans[t + 1].resize(m+1);
 41     for (int i = 1; i <= m; ++i)
 42     {
 43         if (i & 1)ans[t][i] = a[t][i];
 44         else ans[t + 1][i] = a[t][i];
 45     }
 46     ans[t][2] = a[t + 1][m];
 47     ans[t + 1][1] = a[t + 1][m - 1];
 48     for (int i = 3; i <= m; ++i)
 49     {
 50         if (i & 1)ans[t + 1][i] = a[t+1][i-2];
 51         else ans[t][i] = a[t + 1][i-2];
 52     }
 53 }
 54 
 55 void change3(int t)
 56 {
 57     ans[t].resize(m+1);
 58     ans[t + 1].resize(m+1);
 59     ans[t + 2].resize(m+1);
 60     for (int i = 1; i <= m; ++i)
 61     {
 62         if (i & 1)ans[t][i] = a[t][i];
 63         else ans[t + 2][i] = a[t][i];
 64     }
 65     ans[t][2] = a[t + 1][m];
 66     ans[t + 1][1] = a[t + 1][m - 1];
 67     for (int i = 3; i <= m; ++i)
 68     {
 69         if (i & 1)ans[t + 1][i] = a[t + 1][i - 2];
 70         else ans[t][i] = a[t + 1][i - 2];
 71     }
 72     for (int i = 1; i <= m; ++i)
 73     {
 74         if (i & 1)ans[t+2][i] = a[t + 2][i];
 75         else ans[t + 1][i] = a[t + 2][i];
 76     }
 77 }
 78 
 79 void change1(int t)
 80 {
 81     ans[t].push_back(123);
 82     for (int i = 1; i <= m; ++i)
 83     {
 84         if ((i & 1) == cas)
 85             ans[t].push_back(a[t][i]);
 86     }
 87 
 88     for (int i = 1; i <= m; ++i)
 89     {
 90         if ((i & 1) == (cas^1))
 91             ans[t].push_back(a[t][i]);
 92     }
 93 }
 94 
 95 int main()
 96 {
 97     ios::sync_with_stdio(false);
 98     cin.tie(0); cout.tie(0);
 99     int i, j, k;
100     cin >> n >> m;
101     if (n == 3 && m == 3)
102     {
103         cout << "YES" << endl;
104         cout << "1 8 3" << endl;
105         cout << "9 2 4" << endl;
106         cout << "5 7 6" << endl;
107         return 0;
108     }
109     if (n == 1 && m == 1)
110     {
111         cout << "YES" << endl;
112         cout << "1" << endl;
113         return 0;
114     }
115     if (n <= 2 && m <= 2) { cout << "NO" << endl; return 0; }
116     if ((n == 1 && m <= 3) || (m == 1 && n == 3) || (n + m == 5 && n != 1 && m != 1)) { cout << "NO" << endl; return 0; }
117     if (n <= m)
118     {
119         for (int i = 1; i <= n; ++i)
120         {
121             a[i].push_back(123);
122             for (int j = 1; j <= m; ++j)
123                 a[i].push_back((i - 1)*m + j);
124         }
125         if (n != 1)
126         {
127             if (n % 2 == 0)
128             {
129                 for (int i = 1; i <= n; i += 2)
130                     change2(i);
131             }
132             else
133             {
134                 for (int i = 1; i <= n - 3; i += 2)
135                     change2(i);
136                 change3(n - 2);
137             }
138         }
139         else change1(1);
140         cout << "YES" << endl;
141         for (int i = 1; i <= n; ++i)
142         {
143             for (int j = 1; j <= m; ++j)
144                 cout << ans[i][j] << " ";
145             cout << endl;
146         }
147     }
148     else
149     {
150         int cnt = 1;
151         swap(m, n);
152         for (int i = 1; i <= n; ++i)a[i].push_back(23);
153         for (int i = 1; i <= m; ++i)
154             for (int j = 1; j <= n; ++j)
155                 a[j].push_back(cnt++);
156         if (n != 1)
157         {
158             if (n % 2 == 0)
159             {
160                 for (int i = 1; i <= n; i += 2)
161                     change2(i);
162             }
163             else
164             {
165                 for (int i = 1; i <= n - 3; i += 2)
166                     change2(i);
167                 change3(n - 2);
168             }
169         }
170         else change1(1);
171         cout << "YES" << endl;
172         for (int i = 1; i <= m; ++i)
173         {
174             for (int j = 1; j <= n; ++j)
175                 cout << ans[j][i] << " ";
176             cout << endl;
177         }
178     }
179     return 0;
180 }

 

posted @ 2017-12-26 15:31  Meternal  阅读(256)  评论(0)  编辑  收藏  举报