一棵树，每个点有一个民族，和一个人数，求每个子树里最多的民族及其人数，如果一样，输出编号最小的
$n \leq 500000$
sol：
卡莫队的毒瘤题，需要 dsu on tree
大概就是 dfs 顺便维护一个数组叫“当前答案”，每次先把轻儿子加进来，再把重儿子加进来，然后把轻儿子删掉，重儿子继承这个“当前答案”数组
然后由于两点间最多有 log 条重链，复杂度很对劲
#include#define LL long long #define rep(i, s, t) for (register int i = (s), i##end = (t); i <= i##end; ++i) #define dwn(i, s, t) for (register int i = (s), i##end = (t); i >= i##end; --i) using namespace std; inline int read() { int x = 0, f = 1; char ch; for (ch = getchar(); !isdigit(ch); ch = getchar()) if (ch == '-') f = -f; for (; isdigit(ch); ch = getchar()) x = 10 * x + ch - '0'; return x * f; } const int maxn = 400010; int n, m, a[maxn], b[maxn]; int first[maxn], to[maxn << 1], nx[maxn << 1], pnt; inline void add(int u, int v) { to[++pnt] = v; nx[pnt] = first[u]; first[u] = pnt; } int fa[maxn], mxs[maxn], size[maxn]; inline void dfs1(int x) { size[x] = 1; for(int i=first[x];i;i=nx[i]) { if(to[i] == fa[x]) continue; fa[to[i]] = x; dfs1(to[i]); size[x] += size[to[i]]; if(size[to[i]] > size[mxs[x]]) mxs[x] = to[i]; } } int cnt[maxn], now; int ans[maxn], ansn[maxn]; inline void cal(int x, int opt) { cnt[b[x]] += opt * a[x]; if((cnt[b[x]] > cnt[now]) || ((cnt[b[x]] == cnt[now]) && (b[x] < now) && (opt == 1))) now = b[x]; for(int i=first[x];i;i=nx[i]) if(to[i] != fa[x]) cal(to[i], opt); } inline void dfs(int x, int opt) { for(int i=first[x];i;i=nx[i]) if(to[i] != fa[x] && to[i] != mxs[x]) dfs(to[i], 0); if(mxs[x]) dfs(mxs[x], 1); for(int i=first[x];i;i=nx[i]) if(to[i] != fa[x] && to[i] != mxs[x]) cal(to[i], 1); cnt[b[x]] += a[x]; if((cnt[b[x]] > cnt[now]) || ((cnt[b[x]] == cnt[now]) && (b[x] < now))) now = b[x]; ans[x] = now; ansn[x] = cnt[now]; if(!opt) now = 0, cal(x, -1); } int main() { n = read(), m = read(); rep(i, 2, n) { int u = read(), v = read(); add(u, v); add(v, u); } dfs1(1); cnt[0] = -1; rep(i, 1, n) b[i] = read(), a[i] = read(); dfs(1, 1); rep(i, 1, n) printf("%d %d\n", ans[i], ansn[i]); }