8.11 树论+分块

P3320 [SDOI2015] 寻宝游戏

题意相当于是让你动态维护一个点集的最小生成树的边权和×2

一个结论:如果我们让所有节点按照dfn序排序 那么这个最小代价就是2×∑i=1ndis(ui,ui+1)

这里钦定un+1=u1

考虑证明:连通后的子树 每条边在DFS的时候会入一次，又出一次，所以每条边就贡献了两次。而按照 DFS 序相邻取距离就相当于一个 DFS 的过程。

得到结论之后 我们可以用set来维护当前点x的前驱y和后继z 如果是加点 那么答案加上dis(x,y)+dis(x,z)−dis(y,z)

否则我们减去上面的柿子即可

不开long long 0pts 开了就100pts()

#include <bits/stdc++.h>
using namespace std;
#define mid (l+r>>1)
#define endl '\n'
#define inl inline
#define ls p<<1
#define rs p<<1|1
#define lson ls,l,mid
#define rson rs,mid+1,r
#define eb emplace_back
#define pii pair<int,int>
#define int long long
const int N = 2e5 + 5;
char buf[1<<24] , *p1 , *p2;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<24,stdin),p1==p2)?EOF:*p1++)
// #define getchar() cin.get();
int read()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f ;
}

int n , m , vis[N] , dis[N] , ans;
int dep[N] , fa[N] , sz[N] , son[N];
int pos[N] , timer , rev[N] , top[N];

set<int> s;
set<int> :: iterator it;

vector<pii> e[N];
inl void add ( int u , int v , int w ) { e[u].eb(v,w); }

struct LCA
{
	void dfs1 ( int u , int f )
	{
		dep[u] = dep[fa[u] = f] + 1 , sz[u] = 1;
		for ( auto [v,w] : e[u] )
			if ( v ^ f )
			{
				dis[v] = dis[u] + w;
				dfs1 ( v , u );
				sz[u] += sz[v];
				if ( sz[son[u]] < sz[v] ) son[u] = v;
			}	
	}	
	void dfs2 ( int u , int tp )
	{
		top[u] = tp , pos[u] = ++timer , rev[timer] = u;
		if ( son[u] ) dfs2 ( son[u] , tp );
		for ( auto [v,w] : e[u] ) if ( v ^ fa[u] && v ^ son[u] ) dfs2 ( v , v );
	}
	int lca ( int u , int v )
	{
		while ( top[u] != top[v] )
		{
			if ( dep[top[u]] < dep[top[v]] ) swap ( u , v );
			u = fa[top[u]];
		}
		if ( dep[u] < dep[v] ) swap ( u , v );
		return v;
	}
}L;

inl int diss ( int u , int v ) { return dis[u] + dis[v] - 2 * dis[L.lca(u,v)]; }

signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr) , cout.tie(nullptr);
	n = read() , m = read();
	for ( int i = 1 , u , v , w ; i < n ; i ++ ) u = read() , v = read() , w = read() , add ( u , v , w ) , add ( v , u , w );
	L.dfs1 ( 1 , 0 ) , L.dfs2 ( 1 , 1 );
	for ( int i = 1 ; i <= m ; i ++ )
	{
		int x = pos[read()];
		if ( !vis[x] ) s.insert(x);
		int y = rev[ ( it = s.lower_bound(x) ) == s.begin() ? *--s.end() : *--it ];
		int z = rev[ ( it = s.upper_bound(x) ) == s.end() ? *s.begin() : *it ];
		int d = diss ( y , rev[x] ) + diss ( rev[x] , z ) - diss ( y , z );
		if ( vis[x] ) s.erase(x);
		if ( vis[x] ) vis[x] ^= 1 , ans -= d;
		else vis[x] ^= 1 , ans += d;
		cout << ans << endl;
	}
	return 0;
}

P5597 【XR-4】复读

很巧妙的一道题 借用题解中的图:

img

设置运行第一次指令后 机器人走到的点为 u 点 我们称这个点为黑点 那么

因为机器人在根节点运行指令的时候不会跑到根节点上面 所以当机器人在 u 点运行指令的时候 也不会走到 u 往上的点

所以我们的第一次指令必须将 u 外面的所有点都搜掉

我们将 u 外面的所有点都建立成一棵新树即可 那么这些树的叠加就是我们需要的东西

img

除了根到黑点的路径只需要走一遍 别的路径都需要两遍 这样可以算出指令的长度

那么我们 dfs 枚举 u 点 每次钦定它及以上的树作为新树 再往下搜 统计答案即可

建新树时也搜索 需要同时保存在原树上的位置和在新树上的位置

如果原树的结点在新树上没有 新建新树结点即可

如果在新树上走到了黑点(black是为了第一次搜索判断黑点 blacknow是为了其余搜索判断黑点) 就把新树上的位置挪到根

#include <bits/stdc++.h>
using namespace std;
#define mid (l+r>>1)
#define endl '\n'
#define inl inline
#define eb emplace_back
const int N = 2e3 + 5;
const int inf = 0x3f3f3f3f;
char buf[1<<24] , *p1 , *p2;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<24,stdin),p1==p2)?EOF:*p1++)
// #define getchar() cin.get();
int read()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f ;
}

int n , m , rt , tot , totnew , black , blacknew , ans = inf;

struct node { int ls , rs; } t[N] , tnew[N];

void in ( int &u )
{
	char ch = cin.get() - '0';
	u = ++tot;	
	if ( ch & 1 ) in ( t[u].ls );
	if ( ch & 2 ) in ( t[u].rs );
}

void dfs2 ( int u , int v )
{
	if ( u == black || v == blacknew ) { blacknew = v , v = 1; }
	if ( t[u].ls )
	{
		if ( !tnew[v].ls ) tnew[v].ls = ++ totnew;
		dfs2 ( t[u].ls , tnew[v].ls );
	}
	if ( t[u].rs ) 
	{
		if ( !tnew[v].rs ) tnew[v].rs = ++ totnew;
		dfs2 ( t[u].rs , tnew[v].rs );
	}
}

void dfs1 ( int u , int dep )
{
	black = u , blacknew = 0;
	memset ( tnew , 0 , sizeof tnew );
	dfs2 ( 1 , totnew = 1 );
	ans = min ( ans , ( totnew - 1 ) * 2 - dep );
	if ( t[u].ls ) dfs1 ( t[u].ls , dep + 1 );
	if ( t[u].rs ) dfs1 ( t[u].rs , dep + 1 );
}
signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr) , cout.tie(nullptr);
	in(rt);
	dfs1 ( 1 , 0 );
	cout << ans << endl;
	return 0;
}

P4168 [Violet] 蒲公英

暴力出奇迹!正解代码留坑()

需要注意的是正常用b[a[i]]++非常慢 需要暴力遍历编号来统计答案并清空

#include <bits/stdc++.h>
using namespace std;
#define mid (l+r>>1)
#define endl '\n'
#define inl inline
#define eb emplace_back
const int N = 4e4 + 5;
char buf[1<<24] , *p1 , *p2;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<24,stdin),p1==p2)?EOF:*p1++)
// #define getchar() cin.get();
int read()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f ;
}

int n , m , a[N] , b[N] , lsh[N] , lstans , l , r;

signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(0) , cout.tie(0);
	n = read() , m = read();
	for ( int i = 1 ; i <= n ; ++ i ) lsh[i] = a[i] = read();
	sort ( lsh + 1 , lsh + n + 1 );
	int sz = unique ( lsh + 1 , lsh + n + 1 ) - lsh - 1;
	for ( int i = 1 ; i <= n ; ++ i ) a[i] = lower_bound ( lsh + 1 , lsh + sz + 1 , a[i] ) - lsh;
 	for ( int i = 1 ; i <= m ; ++ i )
	{
		l = ( read() + lstans - 1 ) % n + 1 , r = ( read() + lstans - 1 ) % n + 1;
		if ( l > r ) swap ( l , r );
		for ( int j = l ; j <= r ; ++ j ) ++ b[a[j]];
		int ans = 0;
		for ( int j = 1 ; j <= sz ; ++ j ) if ( b[ans] < b[j] ) ans = j;
		for ( int j = 1 ; j <= sz ; ++ j ) b[j] = 0;
		cout << ( lstans = lsh[ans] ) << endl;	
	}
	return 0;
}

upd on 8.13:来补充正解

看到 ai≤109 先上一发离散化

我们可以开两个数组:

  1. f[i][j]表示第i个块到第j个块(编号最小的)众数 枚举i,j块 每次遍历其中的元素并统计即可 复杂度O(n3)
  2. s[i][j]表示前i个块中j出现了几次 我们枚举所有的块 递推s数组 复杂度O(nn)

对于统计 如果两端的差值小于等于1 那么暴力统计

否则为整块中的众数和散块中的众数的整体统计 我们先将所有的散块加入temp统计数组中

答案先设置为整块中的众数 每次枚举散块中的一个数 将temp数组中的这个数的出现次数加上整块中的出现次数 尝试更新众数即可

调了一上午 最大的错误 yy−xx≤1 写成了 xx−yy≤1

#include <bits/stdc++.h>
using namespace std;
#define mid (l+r>>1)
#define endl '\n'
#define inl inline
#define eb emplace_back
const int N = 4e4 + 5;
const int sqrtn = 2e2 + 5;
char buf[1<<24] , *p1 , *p2;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<24,stdin),p1==p2)?EOF:*p1++)
// #define getchar() cin.get();
int read()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f ;
}

int n , m , a[N] , bel[N] , ll[N] , rr[N] , block , tot , lstans , l , r , cnt[N] , s[sqrtn][N] , f[sqrtn][sqrtn];

vector<int> lsh;

void LSH()
{
	for ( int i = 1 ; i <= n ; i ++ ) a[i] = read() , lsh.eb(a[i]);
	sort ( lsh.begin() , lsh.end() );
	lsh.erase ( unique ( lsh.begin() , lsh.end() ) , lsh.end() );
	for ( int i = 1 ; i <= n ; i ++ ) a[i] = lower_bound ( lsh.begin() , lsh.end() , a[i] ) - lsh.begin() + 1;
}

void upd ( int x , int y ) { for ( int i = x ; i <= y ; i ++ ) ++ cnt[a[i]]; }
void clear ( int x , int y ) { for ( int i = x ; i <= y ; i ++ ) cnt[a[i]] = 0; }

void init()
{
	block = (int)sqrt(n) , tot = n / block;
	if ( n % block ) tot ++;
	for ( int i = 1 ; i <= n ; i ++ ) bel[i] = ( i - 1 ) / block + 1;
	for ( int i = 1 ; i <= tot ; i ++ ) ll[i] = ( i - 1 ) * block + 1 , rr[i] = i * block; rr[tot] = n;
	for ( int i = 1 ; i <= tot ; i ++ )
	{
		for ( int j = 1 ; j <= n ; j ++ ) s[i][a[j]] = s[i-1][a[j]];
		for ( int j = ll[i] ; j <= rr[i] ; j ++ ) ++ s[i][a[j]];
	}
	for ( int i = 1 ; i <= tot ; i ++ )
		for ( int j = i ; j <= tot ; j ++ )
		{
			int maxx = f[i][j-1];
			for ( int k = ll[j] ; k <= rr[j] ; k ++ )
				if ( s[j][a[k]] - s[i-1][a[k]] > s[j][maxx] - s[i-1][maxx] || s[j][a[k]] - s[i-1][a[k]] == s[j][maxx] - s[i-1][maxx] && a[k] < maxx ) maxx = a[k];
			f[i][j] = maxx;
		}
}

int query ( int x , int y )
{
	int xx = bel[x] , yy = bel[y] , maxx = 0;
	if ( yy - xx <= 1 ) 
	{
		upd ( x , y );
		for ( int i = x ; i <= y ; i ++ ) if ( cnt[a[i]] > cnt[maxx] || cnt[a[i]] == cnt[maxx] && a[i] < maxx ) maxx = a[i];
		clear ( x , y );
		return lsh[maxx-1];
	}
	upd ( x , rr[xx] ) , upd ( ll[yy] , y );
	maxx = f[xx+1][yy-1];
	for ( int i = x , tmp , pre ; i <= rr[xx] ; i ++ )
	{
		pre = cnt[maxx] + s[yy-1][maxx] - s[xx][maxx] , tmp = cnt[a[i]] + s[yy-1][a[i]] - s[xx][a[i]];
		if ( tmp > pre || tmp == pre && a[i] < maxx ) maxx = a[i];
	}
	for ( int i = ll[yy] , tmp , pre ; i <= y ; i ++ )
	{
		pre = cnt[maxx] + s[yy-1][maxx] - s[xx][maxx] , tmp = cnt[a[i]] + s[yy-1][a[i]] - s[xx][a[i]];
		if ( tmp > pre || tmp == pre && a[i] < maxx ) maxx = a[i];
	}
	clear ( x , rr[xx] ) , clear ( ll[yy] , y );
	return lsh[maxx-1];
}

signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(0) , cout.tie(0);
	n = read() , m = read();
	LSH() , init();
 	for ( int i = 1 ; i <= m ; i ++ )
	{
		l = ( read() + lstans - 1 ) % n + 1 , r = ( read() + lstans - 1 ) % n + 1;
		if ( l > r ) swap ( l , r );
		cout << ( lstans = query ( l , r ) ) << endl;
	}
	return 0;
}

P2801 教主的魔法

第三遍写了......

注意1:二分查找的时候需要查找的是w−add[i]

注意2:查询散块的时候不要忘记加上add[i]

#include <bits/stdc++.h>
using namespace std;
#define mid (l+r>>1)
#define endl '\n'
#define inl inline
#define eb emplace_back
#define int long long 
const int N = 1e6 + 5;
char buf[1<<24] , *p1 , *p2;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<24,stdin),p1==p2)?EOF:*p1++)
#define getchar() cin.get();
int read()
{
	int x = 0 , f = 1;
	char ch = getchar();
	while ( !isdigit(ch) ) { if ( ch == '-' ) f = -1; ch = getchar(); }
	while ( isdigit(ch) ) { x = ( x << 1 ) + ( x << 3 ) + ( ch ^ 48 ); ch = getchar(); }
	return x * f ;
}

int n , q , block , a[N] , b[N] , l[N] , r[N] , bel[N] , tot;
int add[N];

void init()
{
	block = (int)sqrt(n) , tot = n / block;
	if ( n % block ) tot ++;
	for ( int i = 1 ; i <= tot ; i ++ ) l[i] = ( i - 1 ) * block + 1 , r[i] = i * block;
	r[tot] = n;
	for ( int i = 1 ; i <= n ; i ++ ) bel[i] = ( i - 1 ) / block + 1 , b[i] = a[i];
	for ( int i = 1 ; i <= tot ; i ++ ) sort ( b + l[i] , b + r[i] + 1 );
}

void reset ( int x )
{
	for ( int i = l[x] ; i <= r[x] ; i ++ ) b[i] = a[i];
	sort ( b + l[x] , b + r[x] + 1 );
}

void upd ( int x , int y , int w )
{
	int xx = bel[x] , yy = bel[y];
	if ( xx == yy )
	{
		for ( int i = x ; i <= y ; i ++ ) a[i] += w;
		reset(xx); return;
	}
	for ( int i = x ; i <= r[xx] ; i ++ ) a[i] += w;
	for ( int i = l[yy] ; i <= y ; i ++ ) a[i] += w;
	reset(xx) , reset(yy);
	for ( int i = xx + 1 ; i <= yy - 1 ; i ++ ) add[i] += w;
}

int query ( int x , int y , int w )
{
	int xx = bel[x] , yy = bel[y] , res = 0;
	if ( xx == yy )
	{
		for ( int i = x ; i <= y ; i ++ ) res += ( a[i] + add[xx] >= w ); 
		return res;
	}
	for ( int i = x ; i <= r[xx] ; i ++ ) res += ( a[i] + add[xx] >= w );
	for ( int i = l[yy] ; i <= y ; i ++ ) res += ( a[i] + add[yy] >= w );
	for ( int i = xx + 1 ; i <= yy - 1 ; i ++ ) res += r[i] - ( lower_bound ( b + l[i] , b + r[i] + 1 , w - add[i] ) - b - 1 );
	return res;
}

char ch;
signed main ()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr) , cout.tie(nullptr);
	n = read() , q = read();
	for ( int i = 1 ; i <= n ; i ++ ) a[i] = read();
	init();
	for ( int i = 1 ; i <= q ; i ++ )
	{
		cin >> ch;
		int l = read() , r = read() , w = read();
		if ( ch == 'M' ) upd ( l , r , w );
		else cout << query ( l , r , w ) << endl;
	}
	return 0;
}
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