题意
题目链接
sol
树链剖分板子 + 动态开节点线段树板子
#include<bits/stdc++.h> #define pair pair<int, int> #define mp(x, y) make_pair(x, y) #define fi first #define se second //#define int long long #define ll long long #define fin(x) {freopen(#x".in","r",stdin);} #define fout(x) {freopen(#x".out","w",stdout);} using namespace std; const int maxn = 1e6 + 10, mod = 1e9 + 7, inf = 1e9 + 10; const double eps = 1e-9; template <typename a, typename b> inline bool chmin(a &a, b b){if(a > b) {a = b; return 1;} return 0;} template <typename a, typename b> inline bool chmax(a &a, b b){if(a < b) {a = b; return 1;} return 0;} template <typename a, typename b> inline ll add(a x, b y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;} template <typename a, typename b> inline void add2(a &x, b y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);} template <typename a, typename b> inline ll mul(a x, b y) {return 1ll * x * y % mod;} template <typename a, typename b> inline void mul2(a &x, b y) {x = (1ll * x * y % mod + mod) % mod;} template <typename a> inline void debug(a a){cout << a << 'n';} template <typename a> inline ll sqr(a x){return 1ll * x * x;} inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f; } int ch(int x, int l, int r) { return x >= l && x <= r; } int n, q, c[maxn], w[maxn], dep[maxn], fa[maxn], son[maxn], siz[maxn], top[maxn], id[maxn], cnt; vector<int> v[maxn]; void dfs1(int x, int _fa) { dep[x] = dep[_fa] + 1; siz[x] = 1; fa[x] = _fa; for(auto &to : v[x]) { if(to == _fa) continue; dfs1(to, x); siz[x] += siz[to]; if(siz[to] > siz[son[x]]) son[x] = to; } } void dfs2(int x, int topf) { top[x] = topf; id[x] = ++cnt; if(!son[x]) return ; dfs2(son[x], topf); for(auto &to : v[x]) { if(top[to]) continue; dfs2(to, to); } } const int ss = (3e6 + 10); int root[maxn], ls[ss], rs[ss], mx[ss], sum[ss], tot; void update(int k) { mx[k] = max(mx[ls[k]], mx[rs[k]]); sum[k] = sum[ls[k]] + sum[rs[k]]; } void modify(int &k, int l, int r, int p, int v) { if(!k) k = ++tot; if(l == r) {sum[k] = v, mx[k] = v; return ;} int mid = l + r >> 1; if(p <= mid) modify(ls[k], l, mid, p, v); if(p > mid) modify(rs[k], mid + 1, r, p, v); update(k); } int query(int k, int l, int r, int ql, int qr, int opt) { if(ql <= l && r <= qr) return opt == 0 ? mx[k] : sum[k]; int mid = l + r >> 1; if(ql > mid) return query(rs[k], mid + 1, r, ql, qr, opt); else if(qr <= mid) return query(ls[k], l, mid, ql, qr, opt); else return opt == 0 ? max(query(ls[k], l, mid, ql, qr, opt), query(rs[k], mid + 1, r, ql, qr, opt)) : query(ls[k], l, mid, ql, qr, opt) + query(rs[k], mid + 1, r, ql, qr, opt); } int treemax(int x, int y) { int ans = -inf, p = x; while(top[x] != top[y]) { if(dep[top[x]] < dep[top[y]]) swap(x, y); chmax(ans, query(root[c[p]], 1, n, id[top[x]], id[x], 0)); x = fa[top[x]]; } if(dep[x] < dep[y]) swap(x, y); chmax(ans, query(root[c[p]], 1, n, id[y], id[x], 0)); return ans; } int treesum(int x, int y) { int ans = 0, p = x; while(top[x] != top[y]) { if(dep[top[x]] < dep[top[y]]) swap(x, y); ans += query(root[c[p]], 1, n, id[top[x]], id[x], 1); x = fa[top[x]]; } if(dep[x] < dep[y]) swap(x, y); ans += query(root[c[p]], 1, n, id[y], id[x], 1); return ans; } signed main() { n = read(); q = read(); for(int i = 1; i <= n; i++) w[i] = read(), c[i] = read(); for(int i = 1; i <= n - 1; i++) { int x = read(), y = read(); v[x].push_back(y); v[y].push_back(x); } dfs1(1, 0); dfs2(1, 1); for(int i = 1; i <= n; i++) modify(root[c[i]], 1, n, id[i], w[i]); while(q--) { char s[4]; scanf("%s", s); int x = read(), c = read(); if(s[0] == 'c' && s[1] == 'c') { modify(root[c[x]], 1, n, id[x], 0); modify(root[c], 1, n, id[x], w[x]); c[x] = c; } if(s[0] == 'c' && s[1] == 'w') { modify(root[c[x]], 1, n, id[x], c); w[x] = c; } if(s[0] == 'q' && s[1] == 'm') { printf("%dn", treemax(x, c)); } if(s[0] == 'q' && s[1] == 's') { printf("%dn", treesum(x, c)); } } return 0; } /* 5 6 3 1 2 3 1 2 3 3 5 1 1 2 1 3 3 4 3 5 qs 1 5 cc 3 1 qs 1 5 cw 3 3 qs 1 5 qm 2 4 */
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