cf550D. Regular Bridge(構造)

題意

給出一個$k$，構造一個無向圖，使得每個點的度數為$k$，且存在一個橋

Sol

神仙題

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

我簡單的來說一下構造過程

首先$n$是偶數的時候無解

奇數的時候：我們拿出兩個點作為橋

先構建一條橋邊，對於兩個端點分別做同樣操作：

新建$k−1$個點，每個點向端點連邊

再新建$k−1$個點，每個點向相鄰的點連邊

對於兩層點形成的二分圖，兩兩之間連邊

/*
*/
#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 + 10, INF = 1e9 + 10, mod = 1e9 + 7;
const double eps = 1e-9;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int K, N, M;
void Build(int t) {
    for(int i = t + 1; i <= t + K - 1; i++)
        printf("%d %d\n", t, i);
    for(int i = t + 1; i <= t + K - 1; i++)
        for(int j = t + K ; j <= 2 * K + t - 2; j++)
            printf("%d %d\n", i, j);
    for(int i = K + t; i <= 2 * K + t - 2; i++)
        if(((t & 1) && (!(i & 1))) || ((!(t & 1)) && (i & 1))) printf("%d %d\n", i, i + 1);
}
main() {
    K = read();
    if(!(K & 1)) {puts("NO"); return 0;}
    puts("YES");
    N = (2 * (K - 1) + 1) * 2, M = N * K / 2;
    printf("%d %d\n", N, M);
    printf("%d %d\n", 1, N / 2 + 1);
    Build(1); 
    Build(N / 2 + 1);
    return 0;
}
/*
2 2 1
1 1
2 1 1
*/