2023冲刺国赛自测4

A. xor on tree

操作分块，每 $Q/B$ 次遍历整棵树，每个询问需要特殊查询 $B$ 个

复杂度 $\frac{Q}{B}nlog + QB$

大力卡常能过

code

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

int read(){
	int x = 0; char c = getchar();
	while(!isdigit(c))c = getchar();
	do{x = x * 10 + (c ^ 48); c = getchar();}while(isdigit(c));
	return x;
}
const int maxn = 1e5 + 5;
vector<int>son[maxn], point;
int n, w[maxn], q, qnt, ans[maxn], las[maxn];
int tim, dfn[maxn], dfnr[maxn], sum[maxn];
struct node{int tim, x, val;};
struct rem{int qid, tim;};
vector<node>o; vector<rem>qy[maxn]; 
bool in[maxn];
struct Trie{
	int ch[maxn * 31][2], si[maxn * 31], cnt = 1, rt = 1;
	void insert(int x, int op){
		int now = rt;
		for(int i = 30; i >= 0; --i){
			if(!ch[now][(x >> i) & 1])ch[now][(x >> i) & 1] = ++cnt;
			now = ch[now][(x >> i) & 1]; si[now] += op;
		}
	}
	int query(int x){
		int res = 0, now = 1;
		for(int i = 30; i >= 0 && now; --i){
			if(si[ch[now][((x >> i) & 1) ^ 1]])res += (1 << i), now = ch[now][((x >> i) & 1) ^ 1];
			else now = ch[now][(x >> i) & 1];
		}
		return res;
	}
}T;
void pre(int x){
	dfn[x] = ++tim;
	for(int v : son[x])pre(v);
	dfnr[x] = tim;
}
void dfs(int x){
	if(sum[dfn[x] - 1] == sum[dfnr[x]])return;
	if(!in[x])T.insert(w[x], 1);
	for(rem v : qy[x])ans[v.qid] = T.query(v.tim);
	for(int v : son[x])dfs(v);
	if(!in[x])T.insert(w[x], -1);
}
void sol(){
	for(int i = 1; i <= n; ++i)sum[i] += sum[i - 1];
	dfs(1);
	for(node v : o)
		if(v.tim){
			for(int p : point)
				if(dfn[v.x] >= dfn[p] && dfnr[v.x] <= dfnr[p])
					ans[v.val] = max(ans[v.val], w[v.x] ^ w[p]);
		}else w[v.x] = v.val;
	o.clear(); point.clear(); for(int i = 1; i <= n; ++i)sum[i] = in[i] = false, qy[i].clear();
}
int main(){
	freopen("tree.in","r",stdin);
	freopen("tree.out","w",stdout);
	n = read(), q = read();
	for(int i = 1; i <= n; ++i)las[i] = w[i] = read();
	for(int i = 2; i <= n; ++i)son[read()].push_back(i);
	pre(1);
	int B = sqrt(n * 30) * 3.4;
	for(int i = 1; i <= q; ++i){
		int op = read(), u = read();
		if(!op){o.push_back({0, u, read()}); in[u] = true; las[u] = o.back().val; point.push_back(u); }
		else {qy[u].push_back({++qnt, las[u]}); o.push_back({1, u, qnt}); ++sum[dfn[u]];}
		if(i % B == 0)sol();
	}
	sol();
	for(int i = 1; i <= qnt; ++i)printf("%d\n",ans[i]);
	return 0;
}

B. calc on lowbit

转成差分的形式，现在计算$\sum_{i = 1}^r f(i)$

设 $f, g _{i, 1 / 0, 1 / 0}$ 表示考虑后 $i$ 位，是否有进位，是否比 $r$ 大的概率和期望

枚举该位填啥进行转移

code

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

ll read(){
	ll x = 0; char c = getchar();
	while(!isdigit(c))c = getchar();
	do{x = x * 10 + (c ^ 48); c = getchar();}while(isdigit(c));
	return x;
}
const int mod = 998244353, inv2 = (mod + 1) >> 1;

int f[62][2][2], g[62][2][2];
void add(int &x, int y){x += y; if(x >= mod)x -= mod;}
int calc(ll r){
	memset(f, 0, sizeof(f));
	memset(g, 0, sizeof(g));
	g[0][0][0] = 1;
	for(int i = 0; i < 60; ++i){
		for(int j = 0; j <= 1; ++j)
			for(int k = 0; k <= 1; ++k)
				for(int p = 0; p <= 1; ++p){
					bool big = ((r >> i) & 1) < p || (((r >> i) & 1) == p && k); 
					if(j + p == 0){
						add(f[i + 1][0][big], f[i][j][k]);
						add(g[i + 1][0][big], g[i][j][k]);
					}else if(j + p == 1){
						int val = 1ll * (f[i][j][k] + g[i][j][k]) * inv2 % mod;
						int v2 = 1ll * g[i][j][k] * inv2 % mod;
						add(f[i + 1][0][big], val);
						add(g[i + 1][0][big], v2);
						add(f[i + 1][1][big], val);
						add(g[i + 1][1][big], v2);
					}else{
						add(f[i + 1][1][big], f[i][j][k]);
						add(g[i + 1][1][big], g[i][j][k]);
					}
				}
	}
	return (0ll + f[60][0][0] + f[60][1][0] + 2ll * g[60][1][0]) % mod;
}
int main(){
	freopen("calc.in","r",stdin);
	freopen("calc.out","w",stdout);
	int T = read();
	for(int i = 1; i <= T; ++i){
		ll l = read(), r = read();
		printf("%d\n",(calc(r) - calc(l - 1) + mod) % mod);
	}
	return 0;
}

C. color on board

数据范围就很网络流

每个点建出四个点，表示横着涂黑，竖着涂白，没有竖着涂黑，没有横着涂白

对于 $ak + b$ 的 $b$， 钦定在最后的位置贡献

以横黑为例，如果 $(i,j)$ 涂了， $(i + 1, j)$ 没涂，那么需要有 $b$ 的代价

或者其是最后一列，涂白的代价本身就是 $a + b$

其他情况同理

现在考虑单个点的限制

如果最终要涂黑，那么之前不能涂白，改白色的代价为 $inf$， 如果没有横竖涂黑，那么需要单点涂黑，有 $c$ 的代价

建出来长这样

对于白色，可以证明一定不会被同方向不同颜色涂，所以涂横黑必然涂竖白或者单点涂白

并且限制了不能两次涂黑

code

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

int read(){
	int x = 0; char c = getchar();
	while(!isdigit(c))c = getchar();
	do{x = x * 10 + (c ^ 48); c = getchar();}while(isdigit(c));
	return x;
}
typedef long long ll;

const int inf = 0x3f3f3f3f;
const int maxn = 41 * 41 * 5;
const int maxm = maxn * 200;
int n, m, a, b, c;
char mp[45][45];
int hh[45][45], sb[45][45], fsh[45][45], fhb[45][45], cnt;
struct Dinic{
	int s, t, head[maxn], tot = 1;
	struct edge{int to, net, val;}e[maxm];
	void add(int u, int v, int w){
		e[++tot].net = head[u];
		head[u] = tot;
		e[tot].to = v;
		e[tot].val = w;
	}
	void link(int u, int v, int w){add(u, v, w); add(v, u, 0);}
	int dep[maxn], now[maxn];
	bool bfs(){
		for(int i = 1; i <= t; ++i)dep[i] = 0;
		queue<int>q; dep[s] = 1; q.push(s);
		now[s] = head[s];
		while(!q.empty()){
			int x = q.front(); q.pop();
			for(int i = head[x]; i; i = e[i].net){
				int v = e[i].to;
				if(e[i].val > 0 && dep[v] == 0){
					dep[v] = dep[x] + 1;
					now[v] = head[v];
					if(v == t)return true;
					q.push(v);
				}
			}
		}
		return false;
	}
	int dfs(int x, int from){
		if(x == t || from <= 0)return from;
		int res = from, i;
		for(i = now[x]; i; i = e[i].net){
			int v = e[i].to;
			if(e[i].val > 0 && dep[v] == dep[x] + 1){
				int k = dfs(v, min(res, e[i].val));
				if(k <= 0)dep[v] = 0;
				e[i].val -= k;
				e[i ^ 1].val += k;
				res -= k;
				if(res <= 0)break;
			}
		}
		now[x] = i;
		return from - res;
	}
	int dinic(){
		int ans = 0;
		while(bfs())ans += dfs(s, inf);
		return ans;
	}
	void clear(){
		for(int i = 1; i <= t; ++i)head[i] = 0;
		cnt = 0; tot = 1;
	}
	void init(){
		n = read(), m = read(), a = read(), b = read(), c = read();
		for(int i = 1; i <= n; ++i)scanf("%s",mp[i] + 1);
		for(int i = 1; i <= n; ++i)for(int j = 1; j <= m; ++j)hh[i][j] = ++cnt, sb[i][j] = ++cnt, fsh[i][j] = ++cnt, fhb[i][j] = ++cnt;
		s = ++cnt, t = ++cnt;
		for(int i = 1; i <= n; ++i)
			for(int j = 1; j < m; ++j){
				link(hh[i][j + 1], hh[i][j], b);
				link(fhb[i][j], fhb[i][j + 1], b);
			}
		for(int i = 1; i < n; ++i)
			for(int j = 1; j <= m; ++j){
				link(sb[i + 1][j], sb[i][j], b);
				link(fsh[i][j], fsh[i + 1][j], b);
			}
		for(int i = 1; i <= n; ++i)
			for(int j = 1; j <= m; ++j)
				if(mp[i][j] == '#'){
					link(hh[i][j], fsh[i][j], c);
					link(s, sb[i][j], inf);
					link(s, hh[i][j], a + (j == m ? b : 0));
					link(fhb[i][j], t, inf);
					link(fsh[i][j], t, a + (i == n ? b : 0));
				}else{
					link(s, hh[i][j], a + (j == m ? b : 0));
					link(s, sb[i][j], a + (i == n ? b : 0));
					link(fsh[i][j], t, a + (i == n ? b : 0));
					link(fhb[i][j], t, a + (j == m ? b : 0));
					link(sb[i][j], hh[i][j], c);
					link(fsh[i][j], fhb[i][j], c);
					link(fsh[i][j], hh[i][j], inf);
				}
		printf("%d\n", dinic());
		clear();
	}
}w;
int main(){
	freopen("color.in","r",stdin);
	freopen("color.out","w",stdout);
	int T = read();
	for(int i = 1; i <= T; ++i)w.init();
	return 0;
}