hdu5355 Cake(构造)

转载请注明出处： http://www.cnblogs.com/fraud/ ——by fraud

Cake

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1517 Accepted Submission(s): 233
Special Judge

Problem Description

There are

Input

There are multiple test cases. The first line of input contains an integer

Output

For each test case, output "YES" (without the quotes) if it is possible, otherwise output "NO" in the first line.

If it is possible, then output

Sample Input

1 2

5 3

5 2

9 3

Sample Output

YES

1 5

2 1 4

2 2 3

YES

3 1 5 9

3 2 6 7

3 3 4 8

其实这题还是不容易过的，比赛的时候一看居然都过了100人了，而且发现直接贪心是不对的，于是乎乱搞了一发。。。结果居然过了。。。

这题首先只有在不能整除以及n<2m-1的时候才是NO,其余情况都是YES

然后在[2m-1，4m-2]的范围内贪心，然后之后，每2m个相互构一组

  1 /**
  2  * code generated by JHelper
  3  * More info: https://github.com/AlexeyDmitriev/JHelper
  4  * @author xyiyy @https://github.com/xyiyy
  5  */
  6 
  7 #include <iostream>
  8 #include <fstream>
  9 
 10 //#####################
 11 //Author:fraud
 12 //Blog: http://www.cnblogs.com/fraud/
 13 //#####################
 14 //#pragma comment(linker, "/STACK:102400000,102400000")
 15 #include <iostream>
 16 #include <sstream>
 17 #include <ios>
 18 #include <iomanip>
 19 #include <functional>
 20 #include <algorithm>
 21 #include <vector>
 22 #include <string>
 23 #include <list>
 24 #include <queue>
 25 #include <deque>
 26 #include <stack>
 27 #include <set>
 28 #include <map>
 29 #include <cstdio>
 30 #include <cstdlib>
 31 #include <cmath>
 32 #include <cstring>
 33 #include <climits>
 34 #include <cctype>
 35 
 36 using namespace std;
 37 #define pb(X) push_back(X)
 38 #define rep(X, N) for(int X=0;X<N;X++)
 39 #define rep2(X, L, R) for(int X=L;X<=R;X++)
 40 typedef long long ll;
 41 
 42 #define gao() out<<"NO"<<endl
 43 vector<int> ans[11];
 44 int used[50];
 45 ll tot;
 46 
 47 class hdu5355 {
 48 public:
 49     int ave;
 50 
 51     void solve(std::istream &in, std::ostream &out) {
 52         int n, m;
 53         in >> n >> m;
 54         tot = (ll) (n + 1) * n / 2;
 55         if (tot % m != 0) {
 56             gao();
 57             return;
 58         }
 59         int tmp = tot / m;
 60         if (n < 2 * m - 1) {
 61             gao();
 62             return;
 63         }
 64         int num = n / m;
 65         if (tmp == n) {
 66             out << "YES" << endl;
 67             out << 1 << " " << n << endl;
 68             rep(i, m - 1) {
 69                 out << 2 << " " << i + 1 << " " << n - i - 1 << endl;
 70             }
 71             return;
 72         }
 73         if (m == 1) {
 74             out << "YES" << endl;
 75             out << n;
 76             rep2(i, 1, n)out << " " << i;
 77             out << endl;
 78             return;
 79         }
 80         rep2(i, 1, m)ans[i].clear();
 81         int f = n;
 82         int c = 1;
 83         /*while(n>=4*m-1){
 84             rep2(i,1,m){
 85                 if(c&1) ans[i].pb(f - i + 1);
 86                 else ans[i].pb(f - m + i);
 87             }
 88             f -= m;
 89             n -= m;
 90             c++;
 91             rep2(i,1,m){
 92                 if(c&1) ans[i].pb(f - i + 1);
 93                 else ans[i].pb(f - m + i);
 94             }
 95             f -= m;
 96             n -= m;
 97             c++;
 98         }*/
 99         c = (n + 1 - m * 2) % (m * 2) + m * 2 - 1;
100         int d = (n - c) / (2 * m);
101         for (int i = 1, j = c + 1; i <= m; i++) {
102             rep(k, d)ans[i].pb(j++), ans[i].pb(n--);
103         }
104         tot = n;
105         n = c;
106         ave = (ll) (1 + n) * n / 2 / m;
107         set<int> s;
108         rep2(i, 1, n)s.insert(i);
109         rep2(i, 1, m) {
110             rep(j, ave) {
111                 auto it = s.upper_bound(ave - j);
112                 ans[i].pb(*--it);
113                 j += *it;
114                 s.erase(it);
115                 j--;
116             }
117         }
118 
119 
120         // clr(used,0);
121         // rep(i,tot)a[i] = n - i;
122         // if(dfs(0,0,n,1)){
123         //if(dfs(m,n,0,0,n)){
124         out << "YES" << endl;
125         // rep2(i,1,n)ans[used[i]].pb(i);
126         rep2(i, 1, m) {
127             int sz = ans[i].size();
128             out << sz;
129             rep(j, sz) {
130                 out << " " << ans[i][j];
131             }
132             out << endl;
133         }
134         //}else out<<"NO"<<endl;
135         /*if(n % m != 0){
136             gao();
137             return;
138         }
139         if(num&1){
140             if(m&1){
141                 out<<"YES"<<endl;
142                 run(m);
143                 rep2(i,1,m){
144                     out<<num;
145                     rep(j,ans[i].size()){
146                         out<<" "<<ans[i][j];
147                     }
148                     int last = 3 * m + i;
149                     int f = 4 * m;
150                     rep2(j,4,num){
151                         out<<" "<<last;
152                         f += m;
153                         if(j&1)last = f- m + i;
154                         else last = f - i + 1;
155                     }
156                     out<<endl;
157                 }
158 
159             }else{
160                 gao();
161                 return;
162             }
163         }else{
164             out<<"YES"<<endl;
165             rep2(i,1,m){
166                 out<<num;
167                 int last = i;
168                 int f = m;
169                 rep(j,num){
170                     out<<" "<<last;
171                     f += m;
172                     if(j&1)last = f-i+1;
173                     else last = f - m + i;
174                 }
175             }
176         }*/
177 
178     }
179 
180     bool dfs(int num, int now, int u, int m) {
181         if (now == 0) {
182             int i = tot;
183             while (used[i])i--;
184             used[i] = m;
185             if (dfs(num + 1, i, i - 1, m))return 1;
186             used[i] = 0;
187             return 0;
188         }
189         if (now == ave) {
190             if (num == tot)return 1;
191             else return dfs(num, 0, tot, m + 1);
192         }
193         for (int i = u; i > 0; i--) {
194             if (!used[i] && now + i <= ave) {
195                 used[i] = m;
196                 if (dfs(num + 1, now + i, i - 1, m))return 1;
197                 used[i] = 0;
198             }
199         }
200         return false;
201     }
202 
203     bool dfs(int m, int n, int tot, int num, int now) {
204         if (!m) {
205             return 1;
206         }
207         if (tot = ave) {
208             if (dfs(m - 1, n, 0, num, 0))return 1;
209         }
210         if (tot == 0) {
211             int i = n;
212             while (used[i])i--;
213             used[i] = m;
214             if (dfs(m, n, tot + i, num + 1, i))return 1;
215             used[i] = 0;
216             return 0;
217         }
218         for (int i = now; i > 0; i--) {
219             if (!used[i] && tot + i <= ave) {
220                 used[i] = m;
221                 if (dfs(m, n, tot + i, num + 1, i - 1))return 1;
222                 used[i] = 0;
223             }
224         }
225         return false;
226     }
227 
228 };
229 
230 int main() {
231     std::ios::sync_with_stdio(false);
232     std::cin.tie(0);
233     hdu5355 solver;
234     std::istream &in(std::cin);
235     std::ostream &out(std::cout);
236     int n;
237     in >> n;
238     for (int i = 0; i < n; ++i) {
239         solver.solve(in, out);
240     }
241 
242     return 0;
243 }

posted on 2015-08-07 22:04 xyiyy 阅读(646) 评论(2) 编辑收藏举报

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