2021HDU多校6 - 7028 Decomposition(构造)

Posted 2021-09-04 Frozen_Guardian

题目链接：点击查看

题目大意：给出一个 $n$ 个点的无向完全图，现在要求找到 $k$ 条简单路径，每条简单路径的长度符合相应的要求，且每条边只能经过一次

题目保证 $\sum _{i=1}^k{l}_i=\frac{n(n-1)}{2}$

题目分析：变形：2019ICPC(上海) - Spanning Tree Removal

因为题目保证了 $n$ 是奇数，所以多出来的那个点可以当作中介点，将 $\frac{n}{2}$ 条欧拉通路连接成欧拉回路，然后依次输出相应的答案就可以了

具体构造方法大同小异，注意行末空格：

代码：

// Problem: Spanning Tree Removal
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/4370/D
// Memory Limit: 1048576 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast","inline","-ffast-math")
// #pragma GCC target("avx,sse2,sse3,sse4,mmx")
#include<iostream>
#include<cstdio>
#include<string>
#include<ctime>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<climits>
#include<queue>
#include<map>
#include<set>
#include<sstream>
#include<cassert>
#include<bitset>
#include<list>
#include<unordered_map>
#define lowbit(x) (x&-x)
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
template<typename T>
inline void read(T &x)
{
	T f=1;x=0;
	char ch=getchar();
	while(0==isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(0!=isdigit(ch)) x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
	x*=f;
}
template<typename T>
inline void write(T x)
{
	if(x<0){x=~(x-1);putchar('-');}
    if(x>9)write(x/10);
    putchar(x%10+'0');
}
const int inf=0x3f3f3f3f;
const int N=1e6+100;
int main()
{
#ifndef ONLINE_JUDGE
//	freopen("data.in.txt","r",stdin);
//	freopen("data.out.txt","w",stdout);
#endif
//	ios::sync_with_stdio(false);
	int w,kase=0;
	cin>>w;
	while(w--) {
		int n;
		read(n);
		printf("Case #%d: %d\\n",++kase,n/2);
		for(int i=0;i<n/2;i++) {
			int sgn=1,pos=i,pre=-1;
			for(int j=1;j<n;j++) {
				pre=pos;
				pos=(pos+sgn*j+n)%n;
				sgn*=-1;
				printf("%d %d\\n",pre+1,pos+1);
			}
		}
	}
	return 0;
}