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====== 2020/07/11 -- 2020/07/17 周报 ======

===== 团队训练 =====

[[2020-2021:teams:tle233:Niuke01|2020牛客暑期多校第一场]]

[[2020-2021:teams:tle233:Niuke02|2020牛客暑期多校第二场]]

===== Marvolo =====

==== 专题 ====

动态点分治

==== 比赛 ====

[[https://atcoder.jp/contests/aising2020|Atcoder AIsing Programming Contest 2020]]

==== 题目 ====

[[http://cogs.pro:8081/cogs/problem/problem.php?pid=vmSJxVaje|[ZJOI2007 捉迷藏]]]

===== Kevin =====

==== 专题 ====

图论杂项

==== 比赛 ====

无

==== 题目 ====

见本周推荐

===== TownYan =====

==== 专题 ====

无

==== 比赛 ====

[[https://atcoder.jp/contests/aising2020|Atcoder AIsing Programming Contest 2020]]

==== 题目 ====

见本周推荐

===== 本周推荐 =====

==== Marvolo ====

[[http://cogs.pro:8081/cogs/problem/problem.php?pid=vmSJxVaje|[ZJOI2007 捉迷藏]]]

题目大意是给一棵树,然后每个节点会进行黑白染色,求某个时刻,树上最远的两个黑点的距离.

算是动态点分入门题,对于每个点维护两个可删堆来统计答案.每一次修改点的颜色的时候,修改一下点分树中对应的链上每个点的信息就行了.

可删堆的实现比较有意思,是用两个priority_queue封装在一起,一个记录所有元素,一个记录删除了的元素来实现.

==== Kevin ====

[[https://www.lydsy.com/JudgeOnline/problem.php?id=2115|[BZOJ 2115 Xor]]]

**题意**

无向连通图 $N$ 点 $M$ 边，范围 $1e5$，求从 $1$ 走到 $N$ 的最大路径权值异或和（可以不是简单路径）


**题解**

主要考虑环的影响（如果无环则 $1$ 到 $N$ 路径唯一）。实际上所有可能路径的异或和可以等于任意一条 $1$ 到 $N$ 路径的异或和再异或上某些环的异或和（相当于选择岔路）。如果重边和自环也视为环，这个结论也成立。因此通过无向图 tarjan 或者直接 dfs 得到每个环的异或和，求异或线性基，再以随便一条路径异或和 $x$ 为基础，从高位逐个考虑线性基要不要异或上去，得到答案。复杂度 $O(kN+M)$，$k$ 是线性基的常数。

**代码**
<code cpp>
#include <cstdio>
#define ONE 1ull
#define builtin_log2(x) (63-__builtin_clzll(x))

using ULL = unsigned long long;

const int MV(5e4+7), ME(1e5+7), MB(61);

struct Edge
{
	ULL d;
	int next, v;
} ed[2 * ME];
int head[MV], tot;
#define edd(_u, _v, _d) ed[++tot].next=head[_u], ed[tot].v=_v, ed[tot].d=_d, head[_u]=tot

bool vis[MV];
ULL dis[MV];
ULL base[66];

void elimnt(ULL v)
{
	if (v)
		for (int i=builtin_log2(v); i>=0; --i)
		{
			if (v & (ONE << i))
			{
				if (base[i])
					v ^= base[i];
				else
				{
					base[i] = v;
					break;
				}
			}
		}
}

ULL get_max(ULL path_xor)
{
	for (int i=MB; i>=0; --i)
		if ((path_xor ^ base[i]) > path_xor)
			path_xor ^= base[i];

	return path_xor;
}


void dfs(const int u)
{
	for (int i=head[u]; i; i=ed[i].next)
	{
		const int v = ed[i].v;
		if (vis[v])
			elimnt(dis[u] ^ dis[v] ^ ed[i].d);
		else
			vis[v] = true, dis[v] = dis[u] ^ ed[i].d, dfs(v);
	}
}


int main()
{
	int V, E, u, v;
	ULL d;
	scanf("%d %d", &V, &E);
	while (E--)
	{
		scanf("%d %d %llu", &u, &v, &d);
		edd(u, v, d);
		edd(v, u, d);
	}

	vis[1] = true, dfs(1);
	printf("%llu", get_max(dis[V]);

	return 0;
}
</code>


==== TownYan ====

Atcoder::AIsing Programming Contest 2020-F

[[https://atcoder.jp/contests/aising2020/tasks/aising2020_f]]

**题意**

求解一个过程看起来简单但繁琐的计数问题

**题解**

分输入的奇偶分别插值，得到通项

**吐槽**

插值是真的尝试了，分奇偶是真的第五层