题解 P3833 【[SHOI2012]魔法树】

题目

直通车

很显然这是个树刨的板子，树上链查询和子树查询

注意：

1.这个点的树根为 0 而不是 1 所以注意读图时点标号 +1 就解决了

2.注意数据范围\(2^{32}\)

然后板子就能过了

node

#include <iostream>
#include <cstdio>
#define  N  100005
#define int long long
#define lson rt << 1
#define rson rt << 1|1
using namespace std;
int n,m,r;
int pre[N],dep[N],fa[N],son[N],siz[N],top[N],dfn[N],dfn2[N],cnt;
namespace Seg{
	struct Tree{
		int sum,len,lazy;
	}tree[N << 1];
	void push_up(int rt){
		tree[rt].sum = tree[lson].sum + tree[rson].sum;
	}
	void build(int rt,int l,int r){
		tree[rt].len = r - l + 1;
		if(l == r){
		   tree[rt].sum = 0;
			return;
		}
		int mid = (l + r)>>1;
		build(lson, l, mid);
		build(rson, mid + 1, r);
		push_up(rt);
	}
	void pushdown(int rt) {
		if (tree[rt].lazy){
		  tree[lson].sum +=  tree[rt].lazy * tree[lson].len;
		  tree[rson].sum +=  tree[rt].lazy * tree[rson].len;
		  tree[lson].lazy +=  tree[rt].lazy;
		  tree[rson].lazy +=  tree[rt].lazy;
		  tree[rt].lazy = 0;
		}
	}
	void update(int rt,int k,int l,int r,int L,int R){
		if(l >= L && r <= R){
			tree[rt].sum +=  k * tree[rt].len;
			tree[rt].lazy += k;
			return;
		}
		pushdown(rt);
		int mid = (l + r) >> 1;
		if(mid >= L) update(lson, k, l, mid, L, R);
		if(mid < R) update(rson, k, mid + 1, r, L, R);
		push_up(rt);
	}
	int query(int rt,int l,int r,int L,int R){
		if(l >= L&& r <= R) return tree[rt].sum;
		pushdown(rt);
		int mid = (l + r)>>1,ans = 0;
		if(L <= mid) ans += query(lson,l,mid,L,R);
		if(mid < R)  ans += query(rson,mid + 1,r,L,R);
		return ans;
	}
}
namespace Cut {
	struct edge{
		int v,nxt;
	}e[N << 1];
	int head[N],js;
	void add_edge(int u,int v){
     e[++js].v = v;e[js].nxt = head[u];head[u] = js;
	}
	void dfs(int x,int f,int d){
	  dep[x] = d,fa[x] = f,siz[x] = 1;
	  for(int i = head[x]; i;i = e[i].nxt){
	  	   int v = e[i].v;
	  	   if(v != fa[x]){
	  	       dfs(v, x, d + 1);
			   siz[x] += siz[v];
			   if(!son[x]||siz[son[x]] < siz[v])
			   	  son[x] = v;
		   }
	  }
	}
    void dfs2(int x,int tp){
    	dfn[x] =++cnt,pre[cnt] = x,top[x] = tp;
		if(son[x]) dfs2(son[x],tp);
		for(int i = head[x]; i; i = e[i].nxt){
			 int v = e[i].v;
			 if(v != fa[x]&&v != son[x]) dfs2(v,v);
		}
	}
   int query_tree(int x){
   	   return Seg::query(1, 1, n, dfn[x],dfn[x] + siz[x] - 1);
   }
   void change_sum(int x,int y,int k){
		while(top[x] != top[y]){
			if(dep[top[x]] < dep[top[y]]) swap(x,y);
			Seg::update(1, k, 1, n,dfn[top[x]], dfn[x]);
		    x = fa[top[x]];
		}
		if(dep[x] > dep[y]) swap(x,y);
		Seg::update(1, k, 1, n,dfn[x], dfn[y]);
   }
}
signed main() {
	int Q;
	cin>>n;
	for (int i = 1, x, y; i < n; i++) {
	  cin>>x>>y;
	  Cut::add_edge(x + 1, y + 1);
	  Cut::add_edge(y + 1, x + 1);
	}
	Cut::dfs(1, 0, 1);
	Cut::dfs2(1, 1);
	Seg::build(1, 1, n);
	cin >> Q;
	while(Q--){
		int x,y,z;
	  	char opt;
		cin >> opt;
		if (opt == 'A') cin>>x>>y>>z,Cut::change_sum(x + 1, y + 1, z);
		if (opt == 'Q') cin>>x,printf("%lld\n", Cut::query_tree(x + 1));
	}
}