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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;
}
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原文地址：http://www.cnblogs.com/dwtfukgv/p/5907584.html
