参考链接：https://blog.csdn.net/qq_43472263/article/details/104150940
U41492 树上数颜色 https://www.luogu.com.cn/problem/U41492
求子树的颜色数有多少种
/* U41492 树上数颜色 计算v的子树中颜色有多少种 */ #include<bits/stdc++.h> #define rep(i,a,b) for(int i = a;i <= b;i++) #define pb push_back using namespace std; const int N = 100010; int sz[N],son[N];//字树大小、重儿子编号 int col[N],ans[N],now,cnt[N];//i点的颜色，第i个问题的答案，当前颜色数，i颜色的数量 vector<int>G[N]; void dfs(int u,int f){ //找重儿子 sz[u] = 1; for(auto v : G[u]){ if(v == f)continue; dfs(v,u); sz[u] += sz[v]; if(!son[u] || sz[v] > sz[son[u]]){ son[u] = v; } } } void solve(int u,int f,int op){ //暴力更新答案，op = 1为增加答案，op = -1为减少答案 cnt[col[u]] += op; if(op == 1 && cnt[col[u]] == 1)now++; if(op == -1 && cnt[col[u]] == 0)now--; for(auto v : G[u]){ if(v == f)continue; solve(v,u,op); } } void dfs2(int u,int f,int op){ for(auto v : G[u]){ if(v == f)continue; if(son[u] != v)dfs2(v,u,1); //处理轻儿子，消除影响 } if(son[u])dfs2(son[u],u,0); //处理重儿子 cnt[col[u]]++;if(cnt[col[u]] == 1)now++;//加上当前点的贡献 for(auto v : G[u]){ if(v == f)continue; if(son[u] != v)solve(v,u,1);//添加贡献 } ans[u] = now; //更新答案 if(op == 1)solve(u,f,-1);//删除 } int main(){ int n;cin >> n; rep(i,1,n-1){ int u,v;cin >> u >> v; G[u].pb(v);G[v].pb(u); } rep(i,1,n)cin >> col[i]; dfs(1,-1); dfs2(1,-1,0); int q;cin >> q; while(q--){ int x;cin >> x; cout << ans[x] << endl; } return 0; }
求子树中颜色最多的颜色权值之和，相同数量都要加上
#include<map> #include<set> #include<cmath> #include<deque> #include<queue> #include<stack> #include<string> #include<bitset> #include<cstdio> #include<vector> #include<complex> #include<iomanip> #include<cstring> #include<sstream> #include<iostream> #include<algorithm> #include<unordered_map> //#include<bits/stdc++.h> #define fi first #define se second #define eps 1e-8 #define ll long long #define ull unsigned long long #define pb push_back //#define N 101000 #define inf 0x3f3f3f3f #define INF 0x3f3f3f3f3f3f3f3f #define P pair<double, ll> #define pi acos(-1) #define lson l,mid,rt*2 #define rson mid+1,r,rt*2+1 #define lowbit(x) (x&(-x)) #define SZ(x) ((ll)(x).size()) #define met(a,x) memset(a,x,sizeof(a)) #define openin(x) freopen(x, "r", stdin) #define openout(x) freopen(x, "w",stdout) #define rep(i,a,b) for(ll i = a;i <= b;i++) #define bep(i,a,b) for(ll i = a;i >= b;i--) #define ios() ios::sync_with_stdio(0);cin.tie(0);cout.tie(0); using namespace std; ll mod = 1e9+7; ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); } ll lcm(ll a, ll b) { return a * b / gcd(a, b); } int days[15] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 }; int is_leapyear(int y) { return y % 400 == 0 || (y % 4 == 0 && y % 100 != 0); } int dx[8] = {0,0,1,-1,-1,-1,1,1}; int dy[8] = {1,-1,0,0,-1,1,-1,1}; inline ll read() { ll x = 0, sign = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') sign = -1; for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0'; return sign * x; } inline void write(ll x){ char r[30]; int t = 0; if(x < 0){ x = -x; putchar('-'); } do{ r[t++] = x % 10 + '0'; x /= 10; }while(x); while(t)putchar(r[--t]); } inline void writesp(ll x){ write(x); putchar(' '); } inline void writeln(ll x){ write(x); putchar(' '); } inline void writedo(double x,int op){ cout << fixed <<setprecision(op) << x << endl; } ll qpow(ll a, ll b) { ll ans = 1; while (b) { if (b & 1)ans = ans * a % mod; a = a * a % mod; b >>= 1; } return ans; } const int N = 100010; int sz[N],son[N];//字树大小、重儿子编号 ll col[N],ans[N],now,cnt[N],mx; vector<int>G[N]; void dfs(int u,int f){ //找重儿子 sz[u] = 1; for(auto v : G[u]){ if(v == f)continue; dfs(v,u); sz[u] += sz[v]; if(!son[u] || sz[v] > sz[son[u]]){ son[u] = v; } } } void solve(int u,int f,int op){ cnt[col[u]] += op; if(op == 1){ if(cnt[col[u]] > mx){ mx = cnt[col[u]]; now = col[u]; } else if(cnt[col[u]] == mx){ now += col[u]; } } for(auto v : G[u]){ if(v == f)continue; solve(v,u,op); } } void dfs2(int u,int f,int op){ for(auto v : G[u]){ if(v == f)continue; if(son[u] != v)dfs2(v,u,1); //处理轻儿子，消除影响 } if(son[u])dfs2(son[u],u,0); //处理重儿子 cnt[col[u]]++; if(cnt[col[u]] > mx){ mx = cnt[col[u]]; now = col[u]; } else if(cnt[col[u]] == mx){ now += col[u]; } // if(cnt[col[u]] == 1)now++; for(auto v : G[u]){ if(v == f)continue; if(son[u] != v)solve(v,u,1);//添加贡献 } ans[u] = now; if(op == 1)solve(u,f,-1),now=0,mx = 0;//删除 } int main(){ int n;cin >> n; rep(i,1,n)cin >> col[i]; rep(i,1,n-1){ int u,v;cin >> u >> v; G[u].pb(v);G[v].pb(u); } dfs(1,-1); dfs2(1,-1,0); rep(i,1,n)cout << ans[i] << ' '; cout << endl; return 0; }
a子树中深度为b的结点能否构成回文串
#include<map> #include<set> #include<cmath> #include<deque> #include<queue> #include<stack> #include<string> #include<bitset> #include<cstdio> #include<vector> #include<complex> #include<iomanip> #include<cstring> #include<sstream> #include<iostream> #include<algorithm> #include<unordered_map> //#include<bits/stdc++.h> #define fi first #define se second #define eps 1e-8 #define ll long long #define ull unsigned long long #define pb push_back //#define N 101000 #define inf 0x3f3f3f3f #define INF 0x3f3f3f3f3f3f3f3f #define P pair<ll, ll> #define pi acos(-1) #define lson l,mid,rt*2 #define rson mid+1,r,rt*2+1 #define lowbit(x) (x&(-x)) #define SZ(x) ((ll)(x).size()) #define met(a,x) memset(a,x,sizeof(a)) #define openin(x) freopen(x, "r", stdin) #define openout(x) freopen(x, "w",stdout) #define rep(i,a,b) for(ll i = a;i <= b;i++) #define bep(i,a,b) for(ll i = a;i >= b;i--) #define ios() ios::sync_with_stdio(0);cin.tie(0);cout.tie(0); using namespace std; ll mod = 1e9+7; ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); } ll lcm(ll a, ll b) { return a * b / gcd(a, b); } int days[15] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 }; int is_leapyear(int y) { return y % 400 == 0 || (y % 4 == 0 && y % 100 != 0); } int dx[8] = {0,0,1,-1,-1,-1,1,1}; int dy[8] = {1,-1,0,0,-1,1,-1,1}; inline ll read() { ll x = 0, sign = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') sign = -1; for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0'; return sign * x; } inline void write(ll x){ char r[30]; int t = 0; if(x < 0){ x = -x; putchar('-'); } do{ r[t++] = x % 10 + '0'; x /= 10; }while(x); while(t)putchar(r[--t]); } inline void writesp(ll x){ write(x); putchar(' '); } inline void writeln(ll x){ write(x); putchar(' '); } inline void writedo(double x,int op){ cout << fixed <<setprecision(op) << x << endl; } ll qpow(ll a, ll b) { ll ans = 1; while (b) { if (b & 1)ans = ans * a % mod; a = a * a % mod; b >>= 1; } return ans; } const int N = 500010; int sz[N],son[N];//字树大小、重儿子编号 ll bit[N],ans[N],dep[N]; int SON; vector<int>G[N]; vector<P>vec[N]; char s[N]; void dfs(int u,int f){ //找重儿子 sz[u] = 1; if(f != -1)dep[u] = dep[f] + 1; for(auto v : G[u]){ if(v == f)continue; dfs(v,u); sz[u] += sz[v]; if(!son[u] || sz[v] > sz[son[u]]){ son[u] = v; } } } void solve(int u,int f,int op){ bit[dep[u]] ^= (1ll << (s[u-1]-'a')); for(auto v : G[u]){ if(v == f)continue; if(v != SON)solve(v,u,op); } } void dfs2(int u,int f,int op){ for(auto v : G[u]){ if(v == f)continue; if(son[u] != v)dfs2(v,u,1); //处理轻儿子，消除影响 } if(son[u])dfs2(son[u],u,0),SON = son[u]; //处理重儿子 solve(u,f,op); for(auto v : vec[u]){ int cnt = __builtin_popcount(bit[v.second-1]); if(cnt <= 1)ans[v.first] = 1; } SON = 0; if(op == 1)solve(u,f,-1);//删除 } int main(){ // cout << __builtin_popcount(1);//二进制中有多少个1 int n,q;cin >> n >> q; // rep(i,1,n)cin >> col[i]; rep(i,2,n){ int x = read(); G[x].pb(i); } scanf("%s",&s); rep(i,1,q){ int x = read(),y = read(); vec[x].pb(P(i,y)); } dfs(1,-1); dfs2(1,-1,1); rep(i,1,q){ if(ans[i])puts("Yes"); else puts("No"); } return 0; }
a结点的k阶后代有多少个不同的值
#include<map> #include<set> #include<cmath> #include<deque> #include<queue> #include<stack> #include<string> #include<bitset> #include<cstdio> #include<vector> #include<complex> #include<iomanip> #include<cstring> #include<sstream> #include<iostream> #include<algorithm> #include<unordered_map> //#include<bits/stdc++.h> #define fi first #define se second #define eps 1e-8 #define ll long long #define ull unsigned long long #define pb push_back //#define N 101000 #define inf 0x3f3f3f3f #define INF 0x3f3f3f3f3f3f3f3f #define P pair<ll, ll> #define pi acos(-1) #define lson l,mid,rt*2 #define rson mid+1,r,rt*2+1 #define lowbit(x) (x&(-x)) #define SZ(x) ((ll)(x).size()) #define met(a,x) memset(a,x,sizeof(a)) #define openin(x) freopen(x, "r", stdin) #define openout(x) freopen(x, "w",stdout) #define rep(i,a,b) for(ll i = a;i <= b;i++) #define bep(i,a,b) for(ll i = a;i >= b;i--) #define ios() ios::sync_with_stdio(0);cin.tie(0);cout.tie(0); using namespace std; ll mod = 1e9+7; ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); } ll lcm(ll a, ll b) { return a * b / gcd(a, b); } int days[15] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 }; int is_leapyear(int y) { return y % 400 == 0 || (y % 4 == 0 && y % 100 != 0); } int dx[8] = {0,0,1,-1,-1,-1,1,1}; int dy[8] = {1,-1,0,0,-1,1,-1,1}; inline ll read() { ll x = 0, sign = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') sign = -1; for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0'; return sign * x; } inline void write(ll x){ char r[30]; int t = 0; if(x < 0){ x = -x; putchar('-'); } do{ r[t++] = x % 10 + '0'; x /= 10; }while(x); while(t)putchar(r[--t]); } inline void writesp(ll x){ write(x); putchar(' '); } inline void writeln(ll x){ write(x); putchar(' '); } inline void writedo(double x,int op){ cout << fixed <<setprecision(op) << x << endl; } ll qpow(ll a, ll b) { ll ans = 1; while (b) { if (b & 1)ans = ans * a % mod; a = a * a % mod; b >>= 1; } return ans; } const int N = 500010; int sz[N],son[N];//字树大小、重儿子编号 int ans[N],dep[N]; int val[N]; bitset<N>bit; vector<int>G[N]; vector<P>vec[N]; set<int>sta[N]; multiset<int>sta2[N]; void dfs(int u,int f){ //找重儿子 sz[u] = 1; if(f != -1)dep[u] = dep[f] + 1; for(auto v : G[u]){ if(v == f)continue; dfs(v,u); sz[u] += sz[v]; if(!son[u] || sz[v] > sz[son[u]]){ son[u] = v; } } } void solve(int u,int f,int op){ if(op == 1) { sta[dep[u]].insert(val[u]); // sta2[dep[u]].insert(val[u]); } else { sta[dep[u]].erase(val[u]); // sta2[dep[u]].erase(sta2[dep[u]].find(val[u])); // if(sta2[dep[u]].find(val[u]) == sta2[dep[u]].end()){ // sta[dep[u]].erase(val[u]); // } } for(auto v : G[u]){ if(v == f)continue; solve(v,u,op); } } void dfs2(int u,int f,int op){ for(auto v : G[u]){ if(v == f)continue; if(son[u] != v)dfs2(v,u,1); //处理轻儿子，消除影响 } if(son[u])dfs2(son[u],u,0); //处理重儿子 sta[dep[u]].insert(val[u]); // sta2[dep[u]].insert(val[u]); for(auto v : G[u]){ if(son[u] != v)solve(v,u,1); } for(auto v : vec[u]){ ans[v.first] = SZ(sta[dep[u] + v.second]); } if(op == 1)solve(u,f,-1);//删除 } map<string,int>mp; int tot; vector<int>v; int main(){ // cout << __builtin_popcount(b);//二进制中有多少个1 int n = read(); rep(i,1,n){ string str; int pa; cin >> str >> pa; if(!mp.count(str))mp[str] = ++tot; val[i] = mp[str]; if(pa == 0)v.pb(i); else G[pa].pb(i); } // rep(i,1,n){ // cout << val[i] << ' '; // } // cout << endl; int q = read(); rep(i,1,q){ int x = read(),y = read(); vec[x].pb(P(i,y)); } for(auto root : v){ dfs(root,-1); dfs2(root,-1,1); } rep(i,1,q) writeln(ans[i]); return 0; }
与a结点拥有共同k阶祖先的结点有多少个
（倍增找到a的k阶祖先，然后就是上一题CF246E了）
#include<map> #include<set> #include<cmath> #include<deque> #include<queue> #include<stack> #include<string> #include<bitset> #include<cstdio> #include<vector> #include<complex> #include<iomanip> #include<cstring> #include<sstream> #include<iostream> #include<algorithm> #include<unordered_map> //#include<bits/stdc++.h> #define fi first #define se second #define eps 1e-8 #define ll long long #define ull unsigned long long #define pb push_back //#define N 101000 #define inf 0x3f3f3f3f #define INF 0x3f3f3f3f3f3f3f3f #define P pair<ll, ll> #define pi acos(-1) #define lson l,mid,rt*2 #define rson mid+1,r,rt*2+1 #define lowbit(x) (x&(-x)) #define SZ(x) ((ll)(x).size()) #define met(a,x) memset(a,x,sizeof(a)) #define openin(x) freopen(x, "r", stdin) #define openout(x) freopen(x, "w",stdout) #define rep(i,a,b) for(ll i = a;i <= b;i++) #define bep(i,a,b) for(ll i = a;i >= b;i--) #define ios() ios::sync_with_stdio(0);cin.tie(0);cout.tie(0); using namespace std; ll mod = 1e9+7; ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); } ll lcm(ll a, ll b) { return a * b / gcd(a, b); } int days[15] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 }; int is_leapyear(int y) { return y % 400 == 0 || (y % 4 == 0 && y % 100 != 0); } int dx[8] = {0,0,1,-1,-1,-1,1,1}; int dy[8] = {1,-1,0,0,-1,1,-1,1}; inline ll read() { ll x = 0, sign = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') sign = -1; for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0'; return sign * x; } inline void write(ll x){ char r[30]; int t = 0; if(x < 0){ x = -x; putchar('-'); } do{ r[t++] = x % 10 + '0'; x /= 10; }while(x); while(t)putchar(r[--t]); } inline void writesp(ll x){ write(x); putchar(' '); } inline void writeln(ll x){ write(x); putchar(' '); } inline void writedo(double x,int op){ cout << fixed <<setprecision(op) << x << endl; } ll qpow(ll a, ll b) { ll ans = 1; while (b) { if (b & 1)ans = ans * a % mod; a = a * a % mod; b >>= 1; } return ans; } const int N = 100010; int sz[N],son[N];//字树大小、重儿子编号 int ans[N],dep[N]; int val[N]; int pa[N][21]; vector<int>G[N]; vector<P>vec[N]; set<int>sta[N]; void dfs(int u,int f){ //找重儿子 pa[u][0] = f; sz[u] = 1; if(f != -1)dep[u] = dep[f] + 1; for(auto v : G[u]){ if(v == f)continue; dfs(v,u); sz[u] += sz[v]; if(!son[u] || sz[v] > sz[son[u]]){ son[u] = v; } } } void solve(int u,int f,int op){ if(op == 1) { sta[dep[u]].insert(val[u]); } else { sta[dep[u]].erase(val[u]); } for(auto v : G[u]){ if(v == f)continue; solve(v,u,op); } } void dfs2(int u,int f,int op){ for(auto v : G[u]){ if(v == f)continue; if(son[u] != v)dfs2(v,u,1); //处理轻儿子，消除影响 } if(son[u])dfs2(son[u],u,0); //处理重儿子 sta[dep[u]].insert(val[u]); for(auto v : G[u]){ if(son[u] != v)solve(v,u,1); } for(auto v : vec[u]){ ans[v.first] = SZ(sta[dep[v.second]]) - 1; } if(op == 1)solve(u,f,-1);//删除 } int tot; vector<int>v; int Find(int x,int y){ //x的y阶级祖先 int pos = 0; while(y){ if(y & 1)x = pa[x][pos]; if(x == -1)return -1; y >>= 1; pos++; } return x; } int main(){ // cout << __builtin_popcount(b);//二进制中有多少个1 int n = read(); rep(i,1,n){ val[i] = i; int pa; cin >> pa; if(pa == 0)v.pb(i); else G[pa].pb(i); } for(auto root : v){ dfs(root,-1); } for(int i = 1;i <= n;i++){ for(int j = 1;j <= 20;j++){ if(pa[i][j-1] != -1) pa[i][j] = pa[pa[i][j-1]][j-1]; else pa[i][j] = -1; } } int q = read(); rep(i,1,q){ int x = read(),y = read(); int z = Find(x,y); if(z >= 1) vec[z].pb(P(i,x)); // cout << z << ' ' << i << ' ' << x << endl; } for(auto root : v){ dfs2(root,-1,1); } rep(i,1,q) writeln(ans[i]); return 0; }
a子树中，找到层数结点数最多的那一层
#include<map> #include<set> #include<cmath> #include<deque> #include<queue> #include<stack> #include<string> #include<bitset> #include<cstdio> #include<vector> #include<complex> #include<iomanip> #include<cstring> #include<sstream> #include<iostream> #include<algorithm> #include<unordered_map> //#include<bits/stdc++.h> #define fi first #define se second #define eps 1e-8 #define ll long long #define ull unsigned long long #define pb push_back //#define N 101000 #define inf 0x3f3f3f3f #define INF 0x3f3f3f3f3f3f3f3f #define P pair<ll, ll> #define pi acos(-1) #define lson l,mid,rt*2 #define rson mid+1,r,rt*2+1 #define lowbit(x) (x&(-x)) #define SZ(x) ((ll)(x).size()) #define met(a,x) memset(a,x,sizeof(a)) #define openin(x) freopen(x, "r", stdin) #define openout(x) freopen(x, "w",stdout) #define rep(i,a,b) for(ll i = a;i <= b;i++) #define bep(i,a,b) for(ll i = a;i >= b;i--) #define ios() ios::sync_with_stdio(0);cin.tie(0);cout.tie(0); using namespace std; ll mod = 1e9+7; ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); } ll lcm(ll a, ll b) { return a * b / gcd(a, b); } int days[15] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 }; int is_leapyear(int y) { return y % 400 == 0 || (y % 4 == 0 && y % 100 != 0); } int dx[8] = {0,0,1,-1,-1,-1,1,1}; int dy[8] = {1,-1,0,0,-1,1,-1,1}; inline ll read() { ll x = 0, sign = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') sign = -1; for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0'; return sign * x; } inline void write(ll x){ char r[30]; int t = 0; if(x < 0){ x = -x; putchar('-'); } do{ r[t++] = x % 10 + '0'; x /= 10; }while(x); while(t)putchar(r[--t]); } inline void writesp(ll x){ write(x); putchar(' '); } inline void writeln(ll x){ write(x); putchar(' '); } inline void writedo(double x,int op){ cout << fixed <<setprecision(op) << x << endl; } ll qpow(ll a, ll b) { ll ans = 1; while (b) { if (b & 1)ans = ans * a % mod; a = a * a % mod; b >>= 1; } return ans; } const int N = 1000010; int sz[N],son[N];//字树大小、重儿子编号 int ans[N],dep[N],cnt[N],mx,res,mxid; int val[N],last; vector<int>G[N]; void dfs(int u,int f){ //找重儿子 sz[u] = 1; if(f != -1)dep[u] = dep[f] + 1; for(auto v : G[u]){ if(v == f)continue; dfs(v,u); sz[u] += sz[v]; if(!son[u] || sz[v] > sz[son[u]]){ son[u] = v; } } } void solve(int u,int f,int op){ cnt[dep[u]] += op; if(op == 1 && cnt[dep[u]] >= mx){ if(cnt[dep[u]] > mx){ mxid = u; mx = cnt[dep[u]]; } else if(cnt[dep[u]] == mx){ if(dep[u] < dep[mxid]){ mxid = u; } } } for(auto v : G[u]){ if(v == f)continue; solve(v,u,op); } } void dfs2(int u,int f,int op){ for(auto v : G[u]){ if(v == f)continue; if(son[u] != v)dfs2(v,u,1); //处理轻儿子，消除影响 } if(son[u])dfs2(son[u],u,0); //处理重儿子 cnt[dep[u]]++; if(cnt[dep[u]] >= mx){ if(cnt[dep[u]] > mx){ mx = cnt[dep[u]]; mxid = u; } else if(cnt[dep[u]] == mx){ if(dep[u] < dep[mxid]){ mxid = u; } } res = 0; } last = u; for(auto v : G[u]){ if(v == f)continue; if(son[u] != v)solve(v,u,1); } if(mxid)res = dep[mxid] - dep[u]; ans[u] = res; // cout << u << ' ' << mx << ' ' << res << endl; if(op == 1)solve(u,f,-1),res = 0,mx = 0,mxid = 0;//删除 } int main(){ int n = read(); rep(i,1,n-1){ int u = read(),v = read(); G[u].pb(v);G[v].pb(u); } int root = 1; dfs(root,-1); // rep(i,1,n)cout << dep[i] << ' '; // cout << endl; dfs2(root,-1,1); rep(i,1,n) writeln(ans[i]); return 0; }
a子树中出现次数>=k的颜色有多少种
#include<map> #include<set> #include<cmath> #include<deque> #include<queue> #include<stack> #include<string> #include<bitset> #include<cstdio> #include<vector> #include<complex> #include<iomanip> #include<cstring> #include<sstream> #include<iostream> #include<algorithm> #include<unordered_map> //#include<bits/stdc++.h> #define fi first #define se second #define eps 1e-8 #define ll long long #define ull unsigned long long #define pb push_back //#define N 101000 #define inf 0x3f3f3f3f #define INF 0x3f3f3f3f3f3f3f3f #define P pair<ll, ll> #define pi acos(-1) #define lson l,mid,rt*2 #define rson mid+1,r,rt*2+1 #define lowbit(x) (x&(-x)) #define SZ(x) ((ll)(x).size()) #define met(a,x) memset(a,x,sizeof(a)) #define openin(x) freopen(x, "r", stdin) #define openout(x) freopen(x, "w",stdout) #define rep(i,a,b) for(ll i = a;i <= b;i++) #define bep(i,a,b) for(ll i = a;i >= b;i--) #define ios() ios::sync_with_stdio(0);cin.tie(0);cout.tie(0); using namespace std; ll mod = 1e9+7; ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); } ll lcm(ll a, ll b) { return a * b / gcd(a, b); } int days[15] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 }; int is_leapyear(int y) { return y % 400 == 0 || (y % 4 == 0 && y % 100 != 0); } int dx[8] = {0,0,1,-1,-1,-1,1,1}; int dy[8] = {1,-1,0,0,-1,1,-1,1}; inline ll read() { ll x = 0, sign = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') sign = -1; for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0'; return sign * x; } inline void write(ll x){ char r[30]; int t = 0; if(x < 0){ x = -x; putchar('-'); } do{ r[t++] = x % 10 + '0'; x /= 10; }while(x); while(t)putchar(r[--t]); } inline void writesp(ll x){ write(x); putchar(' '); } inline void writeln(ll x){ write(x); putchar(' '); } inline void writedo(double x,int op){ cout << fixed <<setprecision(op) << x << endl; } ll qpow(ll a, ll b) { ll ans = 1; while (b) { if (b & 1)ans = ans * a % mod; a = a * a % mod; b >>= 1; } return ans; } const int N = 100010; int sz[N],son[N];//字树大小、重儿子编号 int ans[N],dep[N],cnt[N],num[N];//答案、深度、cnt[i]代表i颜色出现多少次，num[i]代表出现次数>=i的有多少个 int col[N];//i点的颜色 vector<int>G[N]; vector<P>vec[N]; void dfs(int u,int f){ //找重儿子 sz[u] = 1; if(f != -1)dep[u] = dep[f] + 1; for(auto v : G[u]){ if(v == f)continue; dfs(v,u); sz[u] += sz[v]; if(!son[u] || sz[v] > sz[son[u]]){ son[u] = v; } } } void solve(int u,int f,int op){ if(op == -1){ num[cnt[col[u]]] -= 1; cnt[col[u]] -= 1; } if(op == 1){ cnt[col[u]] += 1; num[cnt[col[u]]] += 1; } for(auto v : G[u]){ if(v == f)continue; solve(v,u,op); } } void dfs2(int u,int f,int op){ for(auto v : G[u]){ if(v == f)continue; if(son[u] != v)dfs2(v,u,1); //处理轻儿子，消除影响 } if(son[u])dfs2(son[u],u,0); //处理重儿子 cnt[col[u]]++;//col[u]出现次数+1 num[cnt[col[u]]]++;//出现cnt[col[u]]次，次数+1 for(auto v : G[u]){ if(v == f)continue; if(son[u] != v)solve(v,u,1); } for(auto v : vec[u]){ ans[v.first] = num[v.second]; } if(op == 1)solve(u,f,-1);//删除 } int main(){ int n = read(),q = read(); rep(i,1,n)col[i] = read(); rep(i,1,n-1){ int u = read(),v = read(); G[u].pb(v);G[v].pb(u); } rep(i,1,q){ int x = read(),y = read(); vec[x].pb(P(i,y)); } int root = 1; dfs(root,-1); dfs2(root,-1,1); rep(i,1,q) writeln(ans[i]); return 0; }
https://codeforces.com/contest/291/problem/E、CF741D Arpa’s letter-marked tree and Mehrdad’s Dokhtar-kosh paths、wannafly Day2 E 阔力梯的树
掌握套路！