牛客网暑期ACM多校训练营（第七场）E：Counting4Cliques（思维）

链接&＃xff1a;https://www.nowcoder.com/acm/contest/145/E

时间限制&＃xff1a;C/C&＃43;&＃43; 1秒&＃xff0c;其他语言2秒
空间限制&＃xff1a;C/C&＃43;&＃43; 262144K&＃xff0c;其他语言524288K
Special Judge, 64bit IO Format: %lld
题目描述
You love doing graph theory problems. You’ve recently stumbled upon a classical problem : Count the number of 4-cliques in an undirected graph.

Given an undirected simple graph G, a 4-clique of G is a set of 4 nodes such that all pairs of nodes in this set are directly connected by an edge.

This task would be too easy for you, wouldn’t it? Thus, your task here is to find an undirected simple graph G with exactly k 4-cliques. Can you solve this task?

输入描述:
The first line of input contains a single integer k(1≤k≤106)k(1≤k≤106).
输出描述:
On the first line, output two space-separated integers, n,m(1≤n≤75,1≤m≤n∗(n−1)/2).n,m(1≤n≤75,1≤m≤n∗(n−1)/2). On the next m lines, output two space-separated integers denoting an edge of the graph u,v(1≤u,v≤n)u,v(1≤u,v≤n), where u and v are the endpoints of the edge.

Your graph must not contain any self-loops or multiple edges between the same pair of nodes. Any graph that has exactly k 4-cliques and satisfies the constraints will be accepted. It can be proven that a solution always exist under the given constraints.
示例1
输入
1
输出
4 6
1 2
1 3
1 4
2 3
2 4
4 3
说明
In the sample, the whole graph is a 4-clique.

题解&＃xff1a;构造一个大小为tt的完全图，和a,b,c,d,e" role="presentation" >a,b,c,d,e$a,b,c,d,e$五个点。
• a,b,c,d,ea,b,c,d,e五个点之间没有边&＃xff0c;他们只会向tt个点连边。
• 如果连了x" role="presentation" >x$x$个&＃xff0c;构成C3xCx3个大小为4的团。
• 找到最大的tt，枚举a&＃xFF0C;b&＃xFF0C;c&＃xFF0C;d&＃xFF0C;" role="presentation" >a，b，c，d，$a，b，c，d，$计算ee。
• 也可以直接背包，并且记录方案。
• 时间复杂度704" role="presentation" >704${70}^4$或者是背包复杂度

#include
using namespace std;
const int MAX&＃61;1e6&＃43;10;
const int MOD&＃61;1e9&＃43;7;
typedef long long ll;
ll c[100][5];
ll v[MAX];
void init()
{c[3][3]&＃61;c[4][4]&＃61;1;for(int i&＃61;5;i<&＃61;80;i&＃43;&＃43;)c[i][4]&＃61;c[i-1][4]*i/(i-4);for(int i&＃61;4;i<&＃61;80;i&＃43;&＃43;)c[i][3]&＃61;c[i-1][3]*i/(i-3);for(int i&＃61;3;i<&＃61;80;i&＃43;&＃43;)v[c[i][3]]&＃61;i;
}
int main()
{init();int cnt;cin>>cnt;int n;for(int i&＃61;1;i<&＃61;70&&c[i][4]<&＃61;cnt;i&＃43;&＃43;)n&＃61;i;cnt-&＃61;c[n][4];for(int i&＃61;0;i<&＃61;n;i&＃43;&＃43;)for(int j&＃61;0;j<&＃61;i;j&＃43;&＃43;)for(int k&＃61;0;k<&＃61;j;k&＃43;&＃43;)for(int h&＃61;0;h<&＃61;k;h&＃43;&＃43;){int res&＃61;cnt-(c[i][3]&＃43;c[j][3]&＃43;c[k][3]&＃43;c[h][3]);if(res<0)break;if(v[res]>n||(res&&v[res]&＃61;&＃61;0))continue;printf("%d %d\n",n&＃43;(i>0)&＃43;(j>0)&＃43;(k>0)&＃43;(h>0)&＃43;(v[res]>0),n*(n-1)/2&＃43;i&＃43;j&＃43;k&＃43;h&＃43;v[res]);for(int x&＃61;1;x<&＃61;n;x&＃43;&＃43;)for(int y&＃61;x&＃43;1;y<&＃61;n;y&＃43;&＃43;)printf("%d %d\n",x,y);if(i)n&＃43;&＃43;;for(int x&＃61;1;x<&＃61;i;x&＃43;&＃43;)printf("%d %d\n",n,x);if(j)n&＃43;&＃43;;for(int x&＃61;1;x<&＃61;j;x&＃43;&＃43;)printf("%d %d\n",n,x);if(k)n&＃43;&＃43;;for(int x&＃61;1;x<&＃61;k;x&＃43;&＃43;)printf("%d %d\n",n,x);if(h)n&＃43;&＃43;;for(int x&＃61;1;x<&＃61;h;x&＃43;&＃43;)printf("%d %d\n",n,x);if(v[res])n&＃43;&＃43;;for(int x&＃61;1;x<&＃61;v[res];x&＃43;&＃43;)printf("%d %d\n",n,x);return 0;}return 0;
}