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Gym 100962G Green Day (找规律)
题意：你用k 个生成树构成一个完全图。
析：n 个点的完全图有n(n-1)/2个边，一个生成树有n-1个边，你有k 个生成树 即边数等于 K(n-1) ，即 n(n-1)/2 == k(n-1) n = 2*k
所以2k 个边足够，你会发现在每个结点只能做一次开头或者结尾。然后找找规律就好。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <cmath>#include <stack>#define print(a) printf("%d\n", (a))#define freopenr freopen("in.txt", "r", stdin)#define freopenw freopen("out.txt", "w", stdout)using namespace std;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const LL LNF = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 1e2 + 5;const int mod = 1e9 + 7;const int dr[] = {-1, 0, 1, 0};const int dc[] = {0, 1, 0, -1};const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};int n, m;const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};inline int Min(int a, int b){ return a < b ? a : b; }inline int Max(int a, int b){ return a > b ? a : b; }inline LL Min(LL a, LL b){ return a < b ? a : b; }inline LL Max(LL a, LL b){ return a > b ? a : b; }inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m;}int main(){ while(scanf("%d", &n) == 1){ m = n << 1; printf("%d\n", m); for (int i = 1; i <= n; ++i){ for (int j = i+1; j <= i + n; ++j) printf("%d %d\n",i, j); for (int j = 1; j <= m-n-1; ++j) printf("%d %d\n",i+n, (i+n+j) % m == 0 ? m : (i+n+j)%m); } } return 0;}
Gym 100962G Green Day (找规律)
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