UOJ 2021 NOI Day2 部分题解

获奖名单

题目传送门

Solution

不难看出，若我们单个 \(x\) 连 \((0,x),(x,0)\)，两个连 \((x,y),(y,x)\) ，除去中间过对称轴的一个两个组，就是找很多个欧拉回路。

直接来就好了。

Code

#include <bits/stdc++.h>
using namespace std;

#define Int register int
#define MAXN 500005

template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}

#define pii pair<int,int>
#define se second
#define fi first

int n,m;
bool vis[MAXN];
vector <pii> g[MAXN];

int cnt,seq[MAXN];
void dfs (int u){
	while (!g[u].empty()){
		pii it = g[u].back();
		g[u].pop_back();
		if (vis[abs(it.se)]) continue;
		vis[abs (it.se)] = 1,dfs (abs (it.fi)),seq[++ cnt] = it.se;
	}
}

signed main(){
	read (n,m);
	for (Int i = 1;i <= n;++ i){
		int opt,x,y;read (opt,x);
		if (opt == 1) g[0].push_back ({x,-i}),g[x].push_back ({0,i});
		else read (y),g[x].push_back ({y,i}),g[y].push_back ({x,-i});
	}
	dfs (0);
	vector <int> ansl,ansr;
	for (Int i = 1;i <= cnt;i += 2) ansl.push_back (seq[i]);
	for (Int i = 2;i <= cnt;i += 2) ansr.push_back (-seq[i]);
	int mid = 0;
	for (Int i = 1;i <= m;++ i){
		sort (g[i].begin(),g[i].end());
		for (Int j = 0;j < g[i].size();j += 2){
			if (g[i][j].fi == i && g[i][j].se == -g[i][j + 1].se){
				mid = abs(g[i][j].se);
				continue;
			}
			if (g[i][j].fi > i || (g[i][j].fi == i && g[i][j].se < 0))
				ansl.push_back (g[i][j].se),
				ansr.push_back (-g[i][j + 1].se);
		}
	}
	for (Int i = ansl.size() - 1;~i;-- i) write (abs(ansl[i])),putchar (' ');
	if (mid) write (mid),putchar (' ');
	for (Int i = 0;i < ansr.size();++ i) write (abs(ansr[i])),putchar (' ');
	putchar ('\n');
	for (Int i = ansl.size() - 1;~i;-- i) write (ansl[i] < 0),putchar (' ');
	if (mid) write (0),putchar (' ');
	for (Int i = 0;i < ansr.size();++ i) write (ansr[i] < 0),putchar (' ');
	putchar ('\n');
	return 0;
}

诡异操作

题目传送门

Solution

考虑使用线段树维护，对于一个线段树上的区间，我们可以维护 \(s_c\) 表示二进制位出现次数二进制第 \(c\) 位为 \(1\) 的和。你发现这个东西可以进行合并，并且取并操作可以直接对于每一个 \(s_c\) 并。

然后对于下取整操作可以直接暴力重构。

复杂度是个米奇妙妙复杂度，据说是 \(\Theta(128n+q\log^2n)\)。

Code

#include <bits/stdc++.h>
using namespace std;

#define Int register int
#define MAXN 300005

template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}

typedef __uint128_t u128;
inline u128 read() {
    static char buf[100];
    scanf("%s", buf);
    // std::cin >> buf;
    u128 res = 0;
    for(int i = 0;buf[i];++i) {
        res = res << 4 | (buf[i] <= '9' ? buf[i] - '0' : buf[i] - 'a' + 10);
    }
    return res;
}
inline void output(u128 res) {
    if(res >= 16)
        output(res / 16);
    putchar(res % 16 >= 10 ? 'a' + res % 16 - 10 : '0' + res % 16);
    //std::cout.put(res % 16 >= 10 ? 'a' + res % 16 - 10 : '0' + res % 16);
}

#define ll u128
ll a[MAXN];
int n,q,len[MAXN << 2];

ll res = ~ll(0);
struct Segment{
	ll val[MAXN << 2][20],any[MAXN << 2],laz[MAXN << 2];
	void pushup (int x){
		memset (val[x],0,sizeof (u128) * len[x]);
		int ls = x << 1,rs = x << 1 | 1;
		val[x][0] = val[ls][0] ^ val[rs][0];ll car = val[ls][0] & val[rs][0];
		for (Int i = 1;i < len[x];++ i){
			ll w = val[ls][i] ^ val[rs][i];
			val[x][i] = w ^ car,car = (car & w) | (val[ls][i] & val[rs][i]);
		}
		val[x][len[x]] = car,any[x] = any[x << 1] | any[x << 1 | 1];
	}
	ll getv (int x){
		ll ans = 0;
		for (Int i = 0;i <= len[x];++ i) ans += val[x][i] << i;
		return ans;
	}
	void pushadd (int x,ll V){
		for (Int i = 0;i <= len[x];++ i) val[x][i] &= V;
		laz[x] &= V,any[x] &= V;
	}
	void pushdown (int x){pushadd (x << 1,laz[x]),pushadd (x << 1 | 1,laz[x]),laz[x] = res;}
	void build (int x,int l,int r){
		laz[x] = res,len[x] = 1;
		while ((1 << len[x]) <= (r - l + 1)) ++ len[x];
		if (l == r) return val[x][0] = any[x] = a[l],void ();
		int mid = l + r >> 1;
		build (x << 1,l,mid),build (x << 1 | 1,mid + 1,r);
		pushup (x);
	}
	void modify1 (int x,int l,int r,int ql,int qr,ll V){
		if (l >= ql && r <= qr) return pushadd (x,V),void ();
		int mid = l + r >> 1;pushdown (x);
		if (ql <= mid) modify1 (x << 1,l,mid,ql,qr,V);
		if (qr > mid) modify1 (x << 1 | 1,mid + 1,r,ql,qr,V);
		pushup (x);
	}
	void div (int x,int l,int r,ll V){
		if (!any[x]) return ;
		if (l == r) return any[x] = (val[x][0] /= V),void ();
		int mid = l + r >> 1;pushdown (x);
		div (x << 1,l,mid,V),div (x << 1 | 1,mid + 1,r,V);
		pushup (x);
	}
	ll query (int x,int l,int r,int ql,int qr){
		if (l >= ql && r <= qr) return getv (x);
		int mid = l + r >> 1;pushdown (x);ll res = 0;
		if (ql <= mid) res += query (x << 1,l,mid,ql,qr);
		if (qr > mid) res += query (x << 1 | 1,mid + 1,r,ql,qr);
		return res;
	}
	void modify2 (int x,int l,int r,int ql,int qr,ll V){
		if (l >= ql && r <= qr) return div (x,l,r,V);
		int mid = l + r >> 1;pushdown (x);
		if (ql <= mid) modify2 (x << 1,l,mid,ql,qr,V);
		if (qr > mid) modify2 (x << 1 | 1,mid + 1,r,ql,qr,V);
		pushup (x);
	}
}T;

signed main(){
	read (n,q);
	for (Int i = 1;i <= n;++ i) a[i] = read ();
	T.build (1,1,n);
	while (q --> 0){
		int opt,qL,qR;ll V;
		read (opt,qL,qR);
		if (opt <= 2){
			V = read ();
			if (opt == 1 && V > 1) T.modify2 (1,1,n,qL,qR,V);
			else if (opt == 2) T.modify1 (1,1,n,qL,qR,V);
		}
		else output (T.query (1,1,n,qL,qR)),putchar ('\n');
	}
	return 0;
}