Codeforces Round #144 (Div. 2) C. Cycles

http://codeforces.com/contest/233/problem/C

题意：求一个具有k个三元环的无向图。

分析：先一个i个节点的无向完全图，其中C(i,3)<=k,剩下k-C(i,3)个三元环未构成，再加j条边（C(j,2)<=未构成的环）直到满足条件。

View Code

 1 /*

 2 Author:Zhaofa Fang

 3 Lang:C++

 4 */

 5 #include <cstdio>

 6 #include <cstdlib>

 7 #include <iostream>

 8 #include <cmath>

 9 #include <cstring>

10 #include <algorithm>

11 #include <string>

12 #include <utility>

13 #include <vector>

14 #include <queue>

15 #include <stack>

16 #include <map>

17 #include <set>

18 using namespace std;

19 typedef long long ll;

20 #define pii pair<int,int>

21 #define pb push_back

22 #define mp make_pair

23 #define fi first

24 #define se second

25 #define lowbit(x) (x&(-x))

26 #define INF (1<<30)

27 

28 int maz[105][105];

29 int main()

30 {

31     #ifndef ONLINE_JUDGE

32     freopen("in","r",stdin);

33     #endif

34     int K;

35     while(cin>>K)

36     {

37         memset(maz,0,sizeof(maz));

38         int i;

39         for(i=3;i*(i-1)*(i-2)/6<=K;i++);

40         i--;

41         K -= i*(i-1)*(i-2)/6;

42         for(int j=0;j<i;j++)

43             for(int k=0;k<i;k++)

44                 maz[j][k] = (j != k);

45         int n=i;

46         while(K)

47         {

48             int j;

49             for(j=2;j*(j-1)/2<=K;j++);

50             j--;

51             for(int k=0;k<j;k++)maz[n][k] = maz[k][n] = 1;

52             K -= (j-1)*j/2;

53             n++;

54         }

55         cout<<n<<endl;

56         for(int x=0;x<n;x++)

57         {

58             for(int y=0;y<n;y++)

59             printf("%d",maz[x][y]);

60             puts("");

61         }

62     }

63     return 0;

64 }