dsu on tree
https://zhuanlan.zhihu.com/p/560661911
https://codeforces.com/contest/600/problem/E
非常裸的题 前面知乎链接里面有讲
const int N = 1e5 + 5;
int n, c[N], id[N], tot = 0;
struct node {
int mx_cnt = 0;//最多的出现次数
ll mx_sum = 0;//出现次数最多的颜色的编号和
map<int, int>cnt;
vector<int>list;
void add(int u) {
cnt[c[u]]++;
if (cnt[c[u]] > mx_cnt)mx_cnt = cnt[c[u]], mx_sum = c[u];
else if (cnt[c[u]] == mx_cnt)mx_sum += c[u];
list.push_back(u);
}
int size() { return list.size(); }
}sub[N];
ll ans[N];
vector<int>g[N];
void dfs(int u, int fa) {
id[u] = ++tot;
int mx_son = -1, mx_sz = 0;
for (int v : g[u]) {
if (v == fa)continue;
dfs(v, u);
if (sub[id[v]].size() > mx_sz) {
mx_sz = sub[id[v]].size();
mx_son = v;
}
}
if(mx_son!=-1)id[u] = id[mx_son];
for (int v : g[u]) {
if (v == fa)continue;
if (v == mx_son)continue;
for (int son : sub[id[v]].list)
sub[id[u]].add(son);
}
sub[id[u]].add(u);
ans[u] = sub[id[u]].mx_sum;
}
void slove() {
cin >> n;
for (int i = 1; i <= n; i++)cin >> c[i];
for (int i = 1; i <= n - 1; i++) {
int u, v; cin >> u >> v;
g[u].push_back(v);
g[v].push_back(u);
}
dfs(1, 0);
for (int i = 1; i <= n; i++)cout << ans[i] << " ";
cout << endl;
}
https://codeforces.com/contest/1009/problem/F
对于树上每个节点 P，求一个最小的 K，使得其子树中到 P距离为K 的节点数最多
分析：
和上面一个题目类似 我们只需要将颜色改为深度 找到深度个数最多的前提下深度最小
#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) x&(-x)
#define ll long long
const int N = 1e6 + 5;
int n, c[N], id[N], tot = 0;
struct node {
int mx_cnt = 0;
int mx_id = 1e9;
map<int, int>cnt;
vector<int>list;
void add(int u) {
cnt[c[u]]++;
if (cnt[c[u]] > mx_cnt)mx_cnt = cnt[c[u]],mx_id=c[u];
else if (cnt[c[u]] == mx_cnt)mx_id=min(c[u],mx_id);
list.push_back(u);
}
int size() { return list.size(); }
}sub[N];
ll ans[N];
vector<int>g[N];
void dfs(int u, int fa) {
id[u] = ++tot;
int mx_son = -1, mx_sz = 0;
for (int i=0;i<g[u].size();i++) {
int v=g[u][i];
if (v == fa)continue;
c[v]=c[u]+1;
dfs(v, u);
if (sub[id[v]].size() > mx_sz) {
mx_sz = sub[id[v]].size();
mx_son = v;
}
}
if(mx_son!=-1)id[u] = id[mx_son];
for (int i=0;i<g[u].size();i++) {
int v=g[u][i];
if (v == mx_son||v==fa)continue;
for(int j=0;j<sub[id[v]].list.size();j++)
sub[id[u]].add(sub[id[v]].list[j]);
}
sub[id[u]].add(u);
ans[u] =sub[id[u]].mx_id-c[u];
}
void solve() {
cin >> n;
for (int i = 1; i <= n - 1; i++) {
int u, v;scanf("%d%d",&u,&v);
g[u].push_back(v);
g[v].push_back(u);
}
dfs(1, 0);
for (int i = 1; i <= n; i++)printf("%d\n",ans[i]);
}
int main(){
solve();
return 0;
}
https://www.luogu.com.cn/problem/U41492
分析：
也是一道很裸的题目
#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) x&(-x)
#define ll long long
const int N = 1e5 + 5;
int n,id[N],c[N] ,tot;
struct node {
int _cnt = 0;
map<int, int>cnt;
vector<int>list;
void add(int u) {
if(!cnt[c[u]])_cnt++;
cnt[c[u]]++;
list.push_back(u);
}
int size() { return list.size(); }
}sub[N];
ll ans[N];
vector<int>g[N];
void dfs(int u, int fa) {
id[u] = ++tot;
int mx_son = -1, mx_sz = 0;
for (int i=0;i<g[u].size();i++) {
int v=g[u][i];
if (v == fa)continue;
dfs(v, u);
if (sub[id[v]].size() > mx_sz) {
mx_sz = sub[id[v]].size();
mx_son = v;
}
}
if(mx_son!=-1)id[u] = id[mx_son];
for (int i=0;i<g[u].size();i++) {
int v=g[u][i];
if (v == mx_son||v==fa)continue;
for(int j=0;j<sub[id[v]].list.size();j++)
sub[id[u]].add(sub[id[v]].list[j]);
}
sub[id[u]].add(u);
ans[u] =sub[id[u]]._cnt;
}
void solve() {
cin >> n;
for (int i = 1; i <= n - 1; i++) {
int u, v;scanf("%d%d",&u,&v);
g[u].push_back(v);
g[v].push_back(u);
}
for (int i = 1; i <= n; i++)scanf("%d",&c[i]);
dfs(1, 0);
int q;
scanf("%d",&q);
while(q--){
int x;scanf("%d",&x);
printf("%d\n",ans[x]);
}
}
int main(){
solve();
return 0;
}
https://codeforces.com/contest/1618/problem/G
const int maxnum = 2e5 + 5;
int a[maxnum], b[maxnum], ans[maxnum];
multiset<int>sta[2 * maxnum], stb[2 * maxnum];
int sum = 0;//答案
struct query { int k, t, id; };//差值，是否为询问，下标
int n, m, q;
struct DSU
{
int fa[2 * maxnum];
void init() { for (int i = 0; i < 2 * maxnum; i++)fa[i] = i; }
int find(int x) { return x == fa[x] ? x : (fa[x] = find(fa[x])); }
void merge(int x, int y) {
x = find(x), y = find(y);
if (sta[x].size() + stb[x].size() < sta[y].size() + stb[y].size())swap(x, y);
for (int it : sta[y])sta[x].insert(it);
for (int it : stb[y])stb[x].insert(it);
while (sta[x].size() && stb[x].size() && *sta[x].begin() < *stb[x].rbegin()) {
int mi = *sta[x].begin(), mx = *stb[x].rbegin();
sum -= mi, sum += mx;
sta[x].erase(sta[x].find(mi)), stb[x].erase(stb[x].find(mx));
sta[x].insert(mx), stb[x].insert(mi);
}
fa[y] = x;
}
} uf;
void slove() {
uf.init();
cin >> n >> m >> q;
vector<pair<int, int>>v; v.push_back({ 0,0 });
for (int i = 1; i <= n; i++) { cin >> a[i]; v.push_back({ a[i],1 }); sum += a[i]; }
for (int i = 1; i <= m; i++) { cin >> b[i]; v.push_back({ b[i],0 }); }
sort(v.begin(), v.end());//一定记得排序，pair的排序是默认按照先first后second
vector<query>que;
//这里将a b 分开
for (int i = 1; i <= n + m; i++) {
if (v[i].second)sta[i].insert(v[i].first);
else stb[i].insert(v[i].first);
}//排序后，保存两数之前的差值
for (int i = 1; i < n + m; i++)que.push_back({ v[i + 1].first - v[i].first,0,i });
for (int i = 1; i <= q; i++) {int qq; cin >> qq;que.push_back({ qq,1,i });}
sort(que.begin(), que.end(), [](const query & a, const query & b) {
if (a.k != b.k)return a.k < b.k;
return a.t < b.t;
});//lambda真好用（bushi
for (auto q : que) {
if (q.t)ans[q.id] = sum;
else uf.merge(q.id, q.id + 1);
}
for (int i = 1; i <= q; i++)cout << ans[i] << endl;
}
https://www.luogu.com.cn/problem/U95602
问题的关键在于 如果维护每个节点的连续段 空间会炸 如果全局变量 各个子树之间又会产生影响
巧妙点：我们只能对轻儿子进行合并操作 所以先更新完重儿子 将关键值now变为dfn[son[x]]即可将亲儿子的连续段合并到重儿子中
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
const int N=200009;
int n,head[N],cnt,p[N],a[N],son[N],siz[N],L[N],R[N],l[N],r[N],Index,DFN[N],rev[N],vis[N],now;
LL ans[N],Ans,s[N];
struct Edge
{
int nxt,to;
}g[N*2];
void add(int from,int to)
{
g[++cnt].nxt=head[from];
g[cnt].to=to;
head[from]=cnt;
}
void init()
{
scanf("%d",&n);
for (int i=2,x;i<=n;i++)
{
scanf("%d",&x);
add(x,i),add(i,x);
}
for (int i=1,x;i<=n;i++)
scanf("%d",&x),a[x]=i;
for (int i=1;i<=n;i++)
scanf("%d",&p[i]);
}
void dfs(int x,int fa)
{
siz[x]=1,DFN[x]=++Index,rev[Index]=x;
for (int i=head[x];i;i=g[i].nxt)
{
int v=g[i].to;
if(v==fa)
continue;
dfs(v,x);
siz[x]+=siz[v];
if(siz[son[x]]<siz[v])
son[x]=v;
}
L[x]=DFN[x],R[x]=Index;
}
LL calc(int x) { return 1ll*x*(x+1)/2; }
void Insert(int x,int k)
{
vis[x]=now,l[x]=r[x]=x;
if(vis[x+1]!=now) l[x+1]=r[x+1]=0;
if(vis[x-1]!=now) l[x-1]=r[x-1]=0;
int xl=0,xr=0;
if(l[x-1]&&r[x+1])
{
xl=x-l[x-1],xr=r[x+1]-x;
r[l[x-1]]=r[x+1],l[r[x+1]]=l[x-1];
}
else if(l[x-1])
{
xl=x-l[x-1];
r[l[x-1]]=x,l[x]=l[x-1];
}
else if(r[x+1])
{
xr=r[x+1]-x;
l[r[x+1]]=x,r[x]=r[x+1];
}
ans[k]=ans[k]-calc(xl)-calc(xr)+calc(xl+xr+1);
}
void DFS(int x,int fa)
{
for (int i=head[x];i;i=g[i].nxt)
{
int v=g[i].to;
if(v==fa||v==son[x])
continue;
DFS(v,x);
}
now=DFN[x];
if(son[x])
{
DFS(son[x],x);
ans[x]+=ans[son[x]];
DFN[x]=now=DFN[son[x]];
for (int i=head[x];i;i=g[i].nxt)
{
int v=g[i].to;
if(v==fa||v==son[x])
continue;
for (int j=L[v];j<=R[v];j++)
Insert(a[rev[j]],x);
}
}
Insert(a[x],x);
LL k=ans[x];
for (int i=head[x];i;i=g[i].nxt)
{
int v=g[i].to;
if(v==fa) continue;
k-=ans[v];
}
Ans+=k*p[x];
}
void work()
{
dfs(1,-1);
DFS(1,-1);
printf("%lld\n",Ans);
}
int main()
{
init();
work();
return 0;
}
https://www.luogu.com.cn/problem/CF570D
如果对每个节点的子集建立一个cnt[26]的map是会超空间的 本题巧妙地讲统计个数变成了异或问题 这样每个节点就只需要开一个map
还有另一种版本 全局变量
原版本code：
#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) x&(-x)
#define ll long long
const int N = 5e5 + 5;
int n,m,id[N],dp[N],tot;
vector<int>pd[N],num[N];
int ans[N];
char c[N];
struct node {
map<int,int>cnt;
vector<int>list;
void add(int u) {
int I=c[u]-'a';
cnt[dp[u]]^=(1<<I);
list.push_back(u);
}
int size() { return list.size(); }
int calc(int d){
int res=0;
for(int i=0;i<26;i++)
if(cnt[d]&(1<<i))res++;
return res;
}
}sub[N];
vector<int>g[N];
void dfs(int u) {
id[u] = ++tot;
int mx_son = -1, mx_sz = 0;
for (int i=0;i<g[u].size();i++) {
int v=g[u][i];
dp[v]=dp[u]+1;
dfs(v);
if (sub[id[v]].size() > mx_sz) {
mx_sz = sub[id[v]].size();
mx_son = v;
}
}
if(mx_son!=-1)id[u] = id[mx_son];
for (int i=0;i<g[u].size();i++) {
int v=g[u][i];
if (v == mx_son)continue;
for(int j=0;j<sub[id[v]].list.size();j++)
sub[id[u]].add(sub[id[v]].list[j]);
}
sub[id[u]].add(u);
for(int i=0;i<pd[u].size();i++){
int ii=num[u][i];
int ak=pd[u][i];
int res=sub[id[u]].calc(ak);
if(res<=1)
ans[ii]=true;
else ans[ii]=false;
}
}
void solve() {
cin >> n>>m;
for (int i = 2; i <= n ; i++) {
int u;scanf("%d",&u);
g[u].push_back(i);
}
for (int i = 1; i <= n; i++)cin>>c[i];
for(int i=1;i<=m;i++){
int x,dep;scanf("%d%d",&x,&dep);
dep--;
pd[x].push_back(dep);
num[x].push_back(i);
}dfs(1);
for(int i=1;i<=m;i++)
if(ans[i])printf("Yes\n");
else printf("No\n");
}
int main(){
solve();
return 0;
}
全局变量版本code：
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#define ll long long
using namespace std;
inline int read(){
int x=0,o=1;char ch=getchar();
while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
if(ch=='-')o=-1,ch=getchar();
while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
return x*o;
}
const int N=500005;
int n,m,visit[N],ans[N];char s[N];
int size[N],son[N],sum[N],val[N],dep[N];
int tot,head[N],nxt[N],to[N];
inline void add(int a,int b){
nxt[++tot]=head[a];head[a]=tot;to[tot]=b;
}
struct node{int h,nxt;}a[N];int Head[N];
inline void Add(int v,int h,int id){
a[id]=(node){h,Head[v]};Head[v]=id;
}
inline void pre_dfs(int u){
size[u]=1;
for(int i=head[u];i;i=nxt[i]){
int v=to[i];dep[v]=dep[u]+1;
pre_dfs(v);size[u]+=size[v];
if(size[v]>size[son[u]])son[u]=v;
}
}
inline void update(int u){
sum[dep[u]]^=val[u];//异或和a
for(int i=head[u];i;i=nxt[i]){
int v=to[i];if(visit[v])continue;
update(v);
}
}
inline bool calc(int x){//计算这个数为1的位有多少个
int cnt=0;
for(int i=0;i<=25;++i)if(x&(1<<i))++cnt;
return cnt<=1;
}
inline void dfs(int u,int keep){
for(int i=head[u];i;i=nxt[i]){
int v=to[i];if(son[u]==v)continue;
dfs(v,0);
}
if(son[u])dfs(son[u],1),visit[son[u]]=1;update(u);
for(int i=Head[u];i;i=a[i].nxt)ans[i]=calc(sum[a[i].h]);
visit[son[u]]=0;if(!keep)update(u);
}
int main(){
n=read(),m=read();
for(int i=2;i<=n;++i){int x=read();add(x,i);}//直接建有向边
scanf("%s",s+1);for(int i=1;i<=n;++i)val[i]=1<<(s[i]-'a');//给节点赋权值
for(int i=1,v,h;i<=m;++i)v=read(),h=read(),Add(v,h,i);//把询问离线
dep[1]=1;//刚开始这里没赋值,调了两个小时
pre_dfs(1);dfs(1,1);for(int i=1;i<=m;++i)puts(ans[i]?"Yes":"No");
return 0;
}
https://www.luogu.com.cn/problem/CF246E
题意：给定一片森林，每次询问一个节点的 K-Son 共有个多少不同的名字。一个节点的 K-Son 即为在该节点子树内的，深度是该节点深度加 K的节点。
分析：
这样子就不能用局部变量 只能用全局变量 下面代码非常标准的全局变量dsu on tree
#include<bits/stdc++.h>
using namespace std;
#define maxn 100010
#define int long long
int n,x,y,m;
string a[maxn];
vector<pair<int,int> >v[maxn];
vector<pair<int,int> >::iterator it;
int head[maxn],Next[maxn],ver[maxn],tot;
void add(int x,int y){
ver[++tot]=y;
Next[tot]=head[x];
head[x]=tot;
}
int dep[maxn],son[maxn],siz[maxn];
void dfs(int x){
siz[x]=1;
for(int i=head[x];i;i=Next[i]){
int y=ver[i];
dep[y]=dep[x]+1;
dfs(y);
siz[x]+=siz[y];
if(siz[son[x]]<siz[y])son[x]=y;
}
}
set<string>S[maxn*2];
int Ans[maxn];
void del(int x){
S[dep[x]].clear();
for(int i=head[x];i;i=Next[i])del(ver[i]);
}
void upd(int x){
S[dep[x]].insert(a[x]);
}
void add(int x){
upd(x);
for(int i=head[x];i;i=Next[i])add(ver[i]);
}
int vis[maxn];
void dsu(int x){
vis[x]=1;
for(int i=head[x];i;i=Next[i])
if(ver[i]!=son[x])dsu(ver[i]),del(ver[i]);
if(son[x])dsu(son[x]);
for(int i=head[x];i;i=Next[i])
if(ver[i]!=son[x])add(ver[i]);
upd(x);
for(it=v[x].begin();it!=v[x].end();it++)Ans[it->second]=S[it->first].size();
}
signed main(){
cin>>n;
for(int i=1;i<=n;i++){
cin>>a[i]>>x;
if(x)add(x,i);
}
for(int i=1;i<=n;i++)
if(!dep[i])dep[i]=1,dfs(i);
cin>>m;
for(int i=1;i<=m;i++)cin>>x>>y,v[x].push_back(make_pair(y+dep[x],i));
for(int i=1;i<=n;i++)if(!vis[i])dsu(i),del(i);
for(int i=1;i<=m;i++)cout<<Ans[i]<<endl;
}
浙公网安备 33010602011771号