cf550D. Regular Bridge(构造)

题意

给出一个$k$，构造一个无向图，使得每个点的度数为$k$，且存在一个桥

Sol

神仙题

一篇写的非常好的博客：http://www.cnblogs.com/mangoyang/p/9302269.html

我简单的来说一下构造过程

首先$n$是偶数的时候无解

奇数的时候：我们拿出两个点作为桥

先构建一条桥边，对于两个端点分别做同样操作：

新建$k−1$个点，每个点向端点连边

再新建$k−1$个点，每个点向相邻的点连边

对于两层点形成的二分图，两两之间连边

/*
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/hash_policy.hpp>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long
#define LL long long
#define ull unsigned long long
#define rg register
#define pt(x) printf("%d ", x);
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
//char obuf[1<<24], *O = obuf;
//void print(int x) {if(x > 9) print(x / 10); *O++ = x % 10 + '0';}
//#define OS  *O++ = ' ';
using namespace std;
//using namespace __gnu_pbds;
const int MAXN = 1e6 + , INF = 1e9 + , mod = 1e9 + ;
const double eps = 1e-;
inline int read() {
    char c = getchar(); int x = , f = ;
    while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}
    while(c >= '' && c <= '') x = x *  + c - '', c = getchar();
    return x * f;
}
int K, N, M;
void Build(int t) {
    for(int i = t + ; i <= t + K - ; i++)
        printf("%d %d\n", t, i);
    for(int i = t + ; i <= t + K - ; i++)
        for(int j = t + K ; j <=  * K + t - ; j++)
            printf("%d %d\n", i, j);
    for(int i = K + t; i <=  * K + t - ; i++)
        if(((t & ) && (!(i & ))) || ((!(t & )) && (i & ))) printf("%d %d\n", i, i + );
}
main() {
    K = read();
    if(!(K & )) {puts("NO"); return ;}
    puts("YES");
    N = ( * (K - ) + ) * , M = N * K / ;
    printf("%d %d\n", N, M);
    printf("%d %d\n", , N /  + );
    Build();
    Build(N /  + );
    return ;
}
/*
2 2 1
1 1
2 1 1
*/