\(sum[u]\) 表示子树 \(dp\) 值的和,\(dp[u]\) 表示子树 \(u\) 的答案,这里我用 \(dp[u][0]\) 表示 \(sum\), \(dp[u][1]\) 表示 \(dp\) 值。考虑以 \(u\) 结点为 \(lca\) 的链,如果不放这条链,答案就是子节点 \(dp\) 值之和,否则答案为链所占的结点的 \(sum\) 之和加上链的权值,但这里重复计算了链上结点的 \(dp\) 值,所以再将链上的 \(dp\) 值减掉。所以开两棵线段树,一棵维护 \(sum\), 一棵维护 \(dp\)。
注意线段树中更新顺序的问题,先更新 \(u\) 结点的 \(sum\) 值,因为在计算 \(u\) 结点 \(dp\) 值时需要用到。处理完该节点所有链后,再更新 \(dp[u]\),因为减掉链上结点 \(dp\) 值的时候不能减去 \(u\) 点 \(dp\) 值。
#include<cstdio> #include<algorithm> #include<cstring> #include<vector> using namespace std; typedef long long ll; const int maxn = 100010; int T, n, m; int dp[maxn][2]; int h[maxn], cnt = 0; struct E{ int to, next; }e[maxn << 1]; void add(int u, int v){ e[++cnt].to = v; e[cnt].next = h[u]; h[u] = cnt; } struct Node{ int sum; }t[2][maxn << 2]; struct Chain{ int u, v, w; }c[maxn]; int dep[maxn], sz[maxn], fa[maxn], son[maxn]; int top[maxn], st[maxn], dfn = 0; vector<Chain> ch[maxn]; // 以 i 为 lca 的链的集合 void dfs1(int u, int par){ dep[u] = dep[par] + 1; sz[u] = 1; fa[u] = par; int maxson=-1; for(int i = h[u] ; i != -1 ; i = e[i].next){ int v = e[i].to; if(v == par) continue; dfs1(v, u); sz[u] += sz[v]; if(sz[v] > maxson){ maxson = sz[v]; son[u] = v; } } } void dfs2(int u, int tp){ top[u] = tp; st[u] = ++dfn; if(!son[u]) return; dfs2(son[u], tp); for(int i = h[u] ; i != -1 ; i = e[i].next){ int v = e[i].to; if(v == fa[u] || v == son[u]) continue; dfs2(v, v); } } void pushup(int i, int P){ t[P][i].sum = t[P][i << 1].sum + t[P][i << 1 | 1].sum; } void build(int i,int l,int r, int P){ if (l==r){ t[P][i].sum = dp[st[l]][P]; return; } int mid = (l + r) >> 1; build(i << 1, l, mid, P); build(i << 1 | 1, mid + 1, r, P); pushup(i, P); } void modify(int i, int k, int l, int r, int p, int P){ if(l == r){ t[P][i].sum = k; return; } int mid = (l + r) >> 1; if(p <= mid) modify(i << 1, k, l, mid, p, P); else modify(i << 1 | 1, k, mid + 1, r, p, P); pushup(i, P); } int query(int i, int l, int r, int x, int y, int P){ if (x <= l && r <= y){ return t[P][i].sum; } int mid = (l + r) >> 1; int ans = 0; if(x <= mid) ans += query(i << 1, l, mid, x, y, P); if(y > mid) ans += query(i << 1 | 1, mid + 1, r, x, y, P); return ans; } int LCA(int u, int v){ while(top[u] != top[v]){ if(dep[top[u]] < dep[top[v]]) swap(u, v); u = fa[top[u]]; } return dep[u] < dep[v] ? u : v; } int qry(int u, int v, int P){ int ans = 0; while(top[u] != top[v]){ if(dep[top[u]] < dep[top[v]]) swap(u, v); ans += query(1, 1, n, st[top[u]], st[u], P); u = fa[top[u]]; } if(dep[u] > dep[v]) swap(u, v); ans += query(1, 1, n, st[u], st[v], P); return ans; } void dfs(int u, int par){ dp[u][0] = dp[u][1] = 0; for(int i = h[u] ; i != -1 ; i = e[i].next){ int v = e[i].to; if(v == par) continue; dfs(v, u); dp[u][0] += dp[v][1]; } dp[u][1] = dp[u][0]; // 不选链 modify(1, dp[u][0], 1, n, st[u], 0); for(int i = 0 ; i < ch[u].size() ; ++i){ // 选链 Chain no = ch[u][i]; dp[u][1] = max(dp[u][1], qry(no.u, no.v, 0) - qry(no.u, no.v, 1) + no.w); } modify(1, dp[u][1], 1, n, st[u], 1); } ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); } return s * f; } int main(){ T = read(); while(T--){ memset(h, -1, sizeof(h)); cnt = 0; memset(son, 0, sizeof(son)); memset(dp, 0, sizeof(dp)); for(int i = 1 ; i <= n ; ++i) ch[i].clear(); dfn = 0; n = read(), m = read(); int u, v; for(int i = 1 ; i <= n - 1 ; ++i){ u = read(), v = read(); add(u, v); add(v, u); } dfs1(1, 0); dfs2(1, 1); for(int i = 1 ; i <= m ; ++i){ c[i].u = read(), c[i].v = read(), c[i].w = read(); int lca = LCA(c[i].u, c[i].v); ch[lca].push_back(c[i]); } build(1, 1, n, 0); build(1, 1, n, 1); dfs(1, 0); printf("%d\n", dp[1][1]); } return 0; }