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;

}

巴特西

Gym 100962G Green Day (找规律)

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