231004校内赛

T1 水管

题解

很简单的一道题，别想复杂了

只要一边 $dfs$ 即可

先将当前点的所有水量给出去，如果缺水就给出去负数

那么等到最后一个节点如果不是刚好合适，那么就把剩余水量回溯回来，无论正负

再如此给下一个连边的点

如果最后一个点刚好合适那么给下一个点的就是 $0$

实现很简单

诅咒不给SPJ的出题人!

#include<bits/stdc++.h>
#define pii pair<int,int>
#define fi first
#define se second
#define N 200010
using namespace std;
int n,m,a[N],sum,f[N<<1],u[N<<1],v[N<<1];
bool vis[N];
vector<pii>g[N];
int dfs(int x,int fl){
	vis[x] = true;
	fl-=a[x];
	int tmp = fl;
	for(pii i : g[x]){
		if(!vis[i.fi]){
			tmp = dfs(i.fi,fl);
			if(u[i.se]==x) f[i.se] = fl-tmp;
			else f[i.se] = -(fl-tmp);
			fl = tmp;
		}
	}
	return fl;
}
int main(){
	freopen("flow.in","r",stdin);
	freopen("flow.out","w",stdout);
	ios::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	cin>>n;
	for(int i = 1;i<=n;i++){
		cin>>a[i];
		sum+=a[i];
	}
	if(sum){
		cout<<"Impossible";
		return 0;
	}
	cin>>m;
	for(int i = 1;i<=m;i++){
		cin>>u[i]>>v[i];
		g[u[i]].push_back({v[i],i});
		g[v[i]].push_back({u[i],i});
	}
	for(int i = 1;i<=n;i++)
		if(!vis[i])
			if(dfs(i,0)){
				cout<<"Impossible";
				return 0;
			}
	cout<<"Possible\n";
	for(int i = 1;i<=m;i++)
		cout<<f[i]<<"\n";
	return 0;
}

T2 宝藏

题解

两种做法，第一种是 $\mathcal{O}(n{m}^2)$ ，第二种 $\mathcal{O}(nm\log m)$

第一种需要处理一个二阶前缀和，然后暴力遍历所有列的区间并用桶存下

每当桶里有值时花费 $\mathcal{O}(n)$ 来用双指针计数处理行对应的个数

如下为第一种代码

#include<bits/stdc++.h>
#define N 100010
#define M 110
#define ll long long
using namespace std;
int n,m,k,suml[N],sumr[M],t[N*M];
ll ans;
ll query(int k){
	ll res = 0,j = 1,l = 1;
	for(int i = 1;i<=n;i++){
		while(suml[i]-suml[l-1]>k&&l<=i) l++;
		while(suml[i]-suml[j-1]>=k&&j<=i) j++;
		res+=j-l;
	}
	return res;
}
int main(){
	freopen("treasure.in","r",stdin);
	freopen("treasure.out","w",stdout);
	ios::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	cin>>n>>m>>k;
	for(int i = 1;i<=n;i++){
		string s;
		cin>>s;
		for(int j = 0;j<m;j++){
			suml[i]+=(s[j]=='$');
			sumr[j+1]+=(s[j]=='$');
		}
	}
	for(int i = 1;i<=n;i++)
		suml[i]+=suml[i-1];
	for(int i = 1;i<=m;i++)
		sumr[i]+=sumr[i-1];
	for(int i = 1;i<=m;i++)
		for(int j = i;j<=m;j++)
			t[sumr[j]-sumr[i-1]]++;
	for(int i = 0;i<=k;i++){
		if(!t[i])continue;
		ans+=query(k-i)*t[i];
	}
	cout<<ans;
	return 0;
}

第二种做法

对于原矩阵维护一个二维前缀和和后缀和，这样宝藏的数目就可以在两个点对上统计贡献

枚举其中一个点对，用数据结构寻找另一个点即可

#include<cstdio>
#define N 100005
#define M 105
#define K 100005
using namespace std;
int n,m,k,cnt,f[N][M],g[N][M],a[N][M];
long long ans;
struct BIT
{
	int v[M];
	void modify(int x)
	{
		if (!x) ++v[0]; 
		for (;x<M && x;x+=x&-x) ++v[x];
	}
	int query(int x)
	{
		int res=0;
		for (;x;x-=x&-x) res+=v[x];
		return res+v[0];
	}
} s[K*3];
int main()
{
	freopen("treasure.in","r",stdin);
	freopen("treasure.out","w",stdout);
	scanf("%d%d%d\n",&n,&m,&k);
	for (int i=1;i<=n;++i,scanf("\n"))
		for (int j=1;j<=m;++j)
			if (getchar()=='$')
				++f[i][j],++g[i][j],++cnt;
	for (int i=1;i<=n;++i) for (int j=1;j<=m;++j)
		f[i][j]+=f[i-1][j]+f[i][j-1]-f[i-1][j-1];
	for (int i=n;i;--i) for (int j=m;j;--j)
		g[i][j]+=g[i+1][j]+g[i][j+1]-g[i+1][j+1];
	for (int i=0;i<=n;++i) for (int j=0;j<=m;++j)
		a[i][j]=f[i][j]-g[i+1][j+1];
	//if (cnt>100000) return printf("?"),0;
	for (int i=1;i<=n;++i)
	{
		for (int j=0;j<m;++j) 
			s[a[i-1][j]+2*cnt].modify(j);
		for (int j=1;j<=m;++j)
			ans+=s[a[i][j]-k+2*cnt].query(j-1);
	}
	printf("%lld",ans);
}

T3 梦色之乡

题解

这题总共有三种做法

第一种，通过点分树来加速找重心的过程从而做到 $\mathcal{O}(n{\log }^{2}n)$ 的复杂度

#include<bits/stdc++.h>
#define pii pair<int,int>
#define fi first
#define se second
#define int long long
#define N 1000010
using namespace std;
int n,m,now,dfn[N],siz[N],sz[N],weight[2],f[N],x,y,cnt,tot,dfncnt,dep[N],fa[N][20],que[N];
bool vis[N];
vector<int>g[N];
unordered_map<int,int>mp[N];
set<pii>s[N];
bool check(int x,int y){//判断x是否在y的子树内 
	return dfn[x]>=dfn[y]&&(dfn[x]<=dfn[y]+siz[y]-1);
}
void dfs(int x,int father){
	siz[x] = 1;
	dfn[x] = ++dfncnt;
	dep[x] = dep[father]+1;
	fa[x][0] = father;
	for(int i = 1;i<=17;i++)
		fa[x][i] = fa[fa[x][i-1]][i-1];
	for(int i : g[x]){
		if(i==father) continue;
		dfs(i,x);
		siz[x]+=siz[i];
		s[x].insert({siz[i],i});
	}
	if(father) s[x].insert({n-siz[x],father});
}
int jump(int x,int y){
	for(int i = 17;i>=0;i--)
		if(dep[fa[x][i]]>=y)
			x = fa[x][i];
	return x;
}
int gt(int p){
	if(check(p,x)||check(p,y))
		return fa[p][0];
	int u = 0,v = 0,newv = 0,r = 0,t = 0,newt = 0;
	if(!check(x,p)){
		u = fa[p][0];
		v = n-siz[p];
		newv = v-siz[x];
	}else{
		u = jump(x,dep[p]+1);
		v = siz[u];
		newv = siz[u]-siz[x];
	}
	if(!check(y,p)){
		r = fa[p][0];
		t = n-siz[p];
		newt = t-siz[y];
	}else{
		r = jump(y,dep[p]+1);
		t = siz[r];
		newt = siz[r]-siz[y];
	}
	if(r==u) newv-=siz[y],r = 0;
	if(u){
		s[p].erase({v,u});
		s[p].insert({newv,u});
	}
	if(r){
		s[p].erase({t,r});
		s[p].insert({newt,r});
	}
	int tmpwei = 0;auto tmp = s[p].end();
	tmp--;
	if((*tmp).fi<=(now/2)){
		weight[tot++] = p;
		if(((*tmp).fi==(now/2))&&(now%2==0))
			weight[tot++] = (*tmp).se;
	}else tmpwei = (*tmp).se;
	if(u){
		s[p].erase({newv,u});
		s[p].insert({v,u});
	}
	if(r){
		s[p].erase({newt,r});
		s[p].insert({t,r});
	}
	return tmpwei;
}
void ddfs(int x,int father){
	sz[x] = 1;f[x] = 0;
	que[++cnt] = x;
	for(int i : g[x]){
		if(i==father||vis[i]) continue;
		ddfs(i,x);
		sz[x]+=sz[i];
		f[x] = max(f[x],sz[i]);
	}
}
int getrt(int x){
	cnt = 0;
	ddfs(x,0);
	int tmp = que[1];
	for(int i = 1;i<=cnt;i++){
		int p = que[i];
		f[p] = max(f[p],cnt-sz[p]);
		if(f[p]<f[tmp]) tmp = p;
	}
	return tmp;
}
int build(int x){
	int u = getrt(x);
	vis[u] = 1;
	for(int i : g[u]){
		if(vis[i]) continue;
		int v = build(i);
		mp[u][i] = v;
	}
	return u;
}
signed main(){
	freopen("dream.in","r",stdin);
	freopen("dream.out","w",stdout);
	ios::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	cin>>n>>m;
	for(int i = 1;i<n;i++){
		int u,v;
		cin>>u>>v;
		g[u].push_back(v);
		g[v].push_back(u);
	}
	dfs(1,0);
	int rt = build(1);
	for(int i = 1;i<=m;i++){
		cin>>x>>y;
		if(dfn[y]<dfn[x]) swap(x,y);
		if(check(y,x)) y = 0;
		if(x==1){
			cout<<"0\n";
			continue;
		}
		now = n-(siz[x]+siz[y]);
		weight[0] = weight[1] = tot = 0;
		int pos = rt;
		while(1){
			int u = gt(pos);
			if(!u) break;
			pos = mp[pos][u];
		}
		if(tot==2&&weight[0]>weight[1])
			swap(weight[0],weight[1]);
		for(int i = 0;i<tot;i++)
			cout<<weight[i]<<" ";
		cout<<"\n";
	}
	return 0;
}

剩下两种做法不是很熟悉，截个图看看就行

#include<bits/stdc++.h>
#define For(i, a, b) for (int i = (a); i <= (b); ++i)
#define Rof(i, a, b) for (int i = (a); i >= (b); --i)
using namespace std;
using i64 = long long;
constexpr int _N = 1e5 + 5;
int n, m, len;
vector<int> e[_N];
mt19937 rnd(114514);
#define rand(l, r) uniform_int_distribution<int>(l, r)(rnd)
int siz[_N], id[_N], dep[_N], dfn, fa[_N];
set<pair<int, int>> fmx[_N];
vector<int> cr;
bool isc[_N];
void Dfs(int x, int F) {
	siz[x] = 1, id[x] = ++dfn, dep[x] = dep[F] + 1, fa[x] = F;
	auto Add = [&x](int s, int siz) {
		fmx[x].insert({ siz, s });
		if (fmx[x].size() > 3)
			fmx[x].erase(fmx[x].begin());
	};
	for (int v : e[x]) if (v ^ F) {
		Dfs(v, x);
		siz[x] += siz[v];
		Add(v, siz[v]);
	}
}
inline bool In(int x, int y) { return id[y] >= id[x] && id[y] < id[x] + siz[x]; }
vector<int> ans;
void Ask(int x, int y) {
	ans.clear();
	if (dep[x] > dep[y]) swap(x, y);
	if (In(x, y)) y = 0;
	int p = 1, tot = n - siz[x] - siz[y];
	auto GetSiz = [](int x, int dx, int dy) {
		if (In(dx, x) || In(dy, x)) return 0;
		int sum = siz[x];
		if (In(x, dx)) sum -= siz[dx];
		if (In(x, dy)) sum -= siz[dy];
		return sum;
	};
	for (int c : cr)
		if (GetSiz(c, x, y) > tot / 2 && dep[c] > dep[p])
			p = c;
	while (tot - GetSiz(p, x, y) <= tot / 2) {
		int nxt = 0, mx = 0;
		for (auto pr : fmx[p])
			if (GetSiz(pr.second, x, y) > mx)
				mx = GetSiz(pr.second, x, y), nxt = pr.second;
		if (mx <= tot / 2) ans.push_back(p);
		p = nxt;
	}
	sort(ans.begin(), ans.end());
	for (int r : ans) cout << r << ' ';
	cout << '\n';
}
signed main() {
	string file = "dream";
	freopen((file + ".in").c_str(), "r", stdin);
	freopen((file + ".out").c_str(), "w", stdout);
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin >> n >> m;
	len = max(1, (int)(sqrt(n) * 1.5));
	For(i, 1, len) isc[rand(1, n)] = 1;
	For(i, 1, n) if (isc[i]) cr.push_back(i);
	For(i, 2, n) {
		int x, y;
		cin >> x >> y;
		e[x].push_back(y);
		e[y].push_back(x);
	}
	Dfs(1, 0);
	For(i, 1, m) {
		int x, y;
		cin >> x >> y;
		if (x == 1 || y == 1) cout << 0 << '\n';
		else Ask(x, y);
	}
}

#include<cstdio>
#include<algorithm>
#include<ctime>
using namespace std;
const int inf=0x3f3f3f3f;
struct node{int size,id,iid;};
struct edge{int last,en,next;} e[200010];
int n,x,y,son[100010],size[100010],dfn[100010],dep[100010];
int st[100010][18],tot,cnt,cs[100010],q,u,v,ccs[100010];
int f[100010][18],val[100010],g[100010][18],leave[100010],ans,w,sum;
void add(int x,int y)
{
	e[++tot].en=y;
	e[tot].next=e[x].last;
	e[x].last=tot;
}
inline int read()
{
	int s=0;
	char ch=getchar();
	while (ch<'0'||ch>'9') ch=getchar();
	while (ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
	return s; 
}
void dfs(int x,int fa)
{
    f[x][0]=fa;dep[x]=dep[fa]+1;size[x]=1;
	dfn[x]=++cnt;son[x]=0;
    for(int i=e[x].last;i;i=e[i].next) 
	    if (e[i].en!=fa)
		{
			dfs(e[i].en,x);
			if (size[son[x]]<size[e[i].en])
			{
				ccs[x]=cs[x];
				cs[x]=son[x];
				son[x]=e[i].en;
			}
			else if (size[cs[x]]<size[e[i].en]) 
			{
				ccs[x]=cs[x];
				cs[x]=e[i].en;
			}
			else if (size[ccs[x]]<size[e[i].en]) ccs[x]=e[i].en;
			size[x]+=size[e[i].en];
		}
	val[x]=size[son[x]]-size[cs[x]];
	if (!cs[x]) val[x]=0x3f3f3f3f; 
	st[x][0]=val[x];g[x][0]=son[x];
    leave[x]=++tot;
}
int lca(int x,int y)
{
	if (dep[x]<dep[y]) swap(x,y);
	for (int i=17;i>=0;i--)
		if (dep[f[x][i]]>=dep[y]) x=f[x][i];
	if (x==y) return x;
	for (int i=17;i>=0;i--)
		if (f[x][i]!=f[y][i]) x=f[x][i],y=f[y][i];
	return f[x][0];
}
int calc(int z,int x)
{
	if (dfn[u]>=dfn[x]&&leave[u]<=leave[x]) z-=size[u];
	if (dfn[v]>=dfn[x]&&leave[v]<=leave[x]) z-=size[v];
	return z;
}
bool getpos(int x,int y)
{
	if (dfn[x]>=dfn[y]&&leave[x]<=leave[y]) return true;
	return false;
}
void sort3(node &a,node &b,node &c)
{
	if (a.size<b.size) swap(a,b);
	if (a.size<c.size) swap(a,c);
	if (b.size<c.size) swap(b,c);
}
void sort3(int &a,int &b,int &c)
{
	if (a<b) swap(a,b);
	if (a<c) swap(a,c);
	if (b<c) swap(b,c);
}
bool get(int x)
{
	int a,b,c;
	a=calc(size[son[x]],son[x]);
	b=calc(size[cs[x]],cs[x]);
	c=calc(size[ccs[x]],ccs[x]);
	sort3(a,b,c);
	if (a<=sum/2) return true;
	return false;
}
void getans(int x)
{
	for (int i=17;i>=0;i--)
		if (f[x][i]&&get(f[x][i])) x=f[x][i];
	int y=0;
	node a,b,c;
	a=(node){calc(size[son[x]],son[x]),getpos(u,son[x])?u:getpos(v,son[x])?v:0,son[x]};
	if (son[x]==u||son[x]==v) a.size=-inf;
	b=(node){calc(size[cs[x]],cs[x]),getpos(u,cs[x])?u:getpos(v,cs[x])?v:0,cs[x]};
	if (son[x]==u||son[x]==v) a.size=-inf;
	c=(node){calc(size[ccs[x]],ccs[x]),getpos(u,ccs[x])?u:getpos(v,ccs[x])?v:0,ccs[x]};
	if (son[x]==u||son[x]==v) a.size=-inf;
	sort3(a,b,c);
	if (sum-a.size<=sum/2&&a.size!=-inf) y=a.iid;
	if (x>y) swap(x,y);
	if (x) printf("%d %d\n",x,y);
	else printf("%d\n",y);
}
void solve(int x)
{
	bool bj=getpos(w,x);
	if (!bj)
	{
		for (int i=17;i>=0;i--)
			if (g[x][i]) x=g[x][i];
		getans(x);
		return;
 	} 
	else
	{	
		for (int i=17;i>=0;i--)
		{
			int y=g[x][i];
			if (!y||!getpos(w,y)&&y!=w||getpos(y,u)||getpos(y,v)) continue;
			if (calc(st[x][i],y)>=0) x=y;
		}
	}
	node a,b,c;
	if (x==u||x==v) x=f[x][0];
	a=(node){calc(size[son[x]],son[x]),getpos(u,son[x])?u:getpos(v,son[x])?v:0,son[x]};
	if (son[x]==u||son[x]==v||!son[x]) a.size=-inf;
	b=(node){calc(size[cs[x]],cs[x]),getpos(u,cs[x])?u:getpos(v,cs[x])?v:0,cs[x]};
	if (cs[x]==u||cs[x]==v||!cs[x]) b.size=-inf;
	c=(node){calc(size[ccs[x]],ccs[x]),getpos(u,ccs[x])?u:getpos(v,ccs[x])?v:0,ccs[x]};
	if (ccs[x]==u||ccs[x]==v||!ccs[x]) c.size=-inf;
	sort3(a,b,c);
	if (a.size==-inf)
	{
		getans(x);
		return;
	}
	if (getpos(a.iid,w)) w=a.id;
	solve(a.iid);
}
int main()
{
	freopen("dream.in","r",stdin);
	freopen("dream.out","w",stdout);
	n=read();q=read(); 
	for (int i=1;i<n;i++)
		x=read(),y=read(),add(x,y),add(y,x),st[i][0]=inf;
	st[0][0]=inf;	
	dfs(1,0);
	for (int j=1;j<=17;j++)
		for (int i=1;i<=n;i++)
		{
			f[i][j]=f[f[i][j-1]][j-1],g[i][j]=g[g[i][j-1]][j-1];
			st[i][j]=min(st[i][j-1],st[g[i][j-1]][j-1]);
		}
	dfn[0]=leave[0]=size[0]=0;
	while (q--)
	{
		
		scanf("%d%d",&u,&v);ans=0;
		if (u==1||v==1)
		{
			printf("0\n");
			continue;
		}
		w=lca(u,v);
		if (u==w) v=0;
		else if (v==w) u=0;
		sum=n-size[u]-size[v];
		solve(1);
	}
	return 0;
}

T4

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