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  • Bzoj3730: 震波

    题面

    戳我

    Sol

    动态点分治:
    建个点分树,每个节点开两颗线段树,以与该点的距离为下标,维护价值和
    一棵树维护这个点的,一棵维护对上层重心的贡献

    然后。。
    然后?直接暴力搞就行了

    注意常数优化我TLE了一遍,第二遍卡过去的

    # include <bits/stdc++.h>
    # define RG register
    # define IL inline
    # define Fill(a, b) memset(a, b, sizeof(a))
    using namespace std;
    typedef long long ll;
    const int _(2e5 + 10), __(2e7 + 10);
    
    IL ll Read(){
    	RG ll x = 0, z = 1; RG char c = getchar();
    	for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    	for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    	return x * z;
    }
    
    int n, fst[_], nxt[_], w[_], to[_], cnt, Q;
    
    IL void Add(RG int u, RG int v){  nxt[cnt] = fst[u]; to[cnt] = v; fst[u] = cnt++;  }
    
    namespace ChainDiv{
    	int fa[_], size[_], top[_], deep[_], son[_], dfn[_], Index;
    
    	IL void Dfs1(RG int u){
    		size[u] = 1;
    		for(RG int e = fst[u]; e != -1; e = nxt[e]){
    			if(size[to[e]]) continue;
    			deep[to[e]] = deep[u] + 1; fa[to[e]] = u;
    			Dfs1(to[e]);
    			size[u] += size[to[e]];
    			if(size[to[e]] > size[son[u]]) son[u] = to[e];
    		}
    	}
    
    	IL void Dfs2(RG int u, RG int Top){
    		top[u] = Top; dfn[u] = ++Index;
    		if(son[u]) Dfs2(son[u], Top);
    		for(RG int e = fst[u]; e != -1; e = nxt[e]) if(!dfn[to[e]]) Dfs2(to[e], to[e]);
    	}
    
    	IL ll Dis(RG int u, RG int v){
    		RG ll dis = deep[u] + deep[v];
    		while(top[u] ^ top[v]){  if(deep[top[u]] < deep[top[v]]) swap(u, v); u = fa[top[u]];  }
    		if(deep[u] > deep[v]) swap(u, v);
    		return dis - 2 * deep[u];
    	}
    }
    
    int rot[_], ls[__], rs[__], sum[__], tot;
    int size[_], mx[_], frt[_], vis[_], rt, sz, root;
    
    IL void Getroot(RG int u, RG int ff){
    	size[u] = 1; mx[u] = 0;
    	for(RG int e = fst[u]; e != -1; e = nxt[e]){
    		if(vis[to[e]] || to[e] == ff) continue;
    		Getroot(to[e], u);
    		size[u] += size[to[e]];
    		mx[u] = max(mx[u], size[to[e]]);
    	}
    	mx[u] = max(mx[u], sz - size[u]);
    	if(mx[u] < mx[rt]) rt = u;
    }
    
    IL void Create(RG int u, RG int ff){
    	frt[u] = ff; vis[u] = 1;
    	for(RG int e = fst[u]; e != -1; e = nxt[e]){
    		if(vis[to[e]]) continue;
    		rt = 0; sz = size[to[e]];
    		Getroot(to[e], u); Create(rt, u);
    	}
    }
    
    IL void Update(RG int &x, RG int l, RG int r, RG int pos, RG int d){
    	if(!x) x = ++tot; sum[x] += d;
    	if(l == r) return;
    	RG int mid = (l + r) >> 1;
    	if(pos <= mid) Update(ls[x], l, mid, pos, d);
    	else Update(rs[x], mid + 1, r, pos, d);
    }
    
    IL int Sum(RG int x, RG int l, RG int r, RG int k){
    	if(!x) return 0;
    	if(k >= r) return sum[x];
    	RG int mid = (l + r) >> 1;
    	if(k > mid) return sum[ls[x]] + Sum(rs[x], mid + 1, r, k);
    	return Sum(ls[x], l, mid, k);
    }
    
    IL void Modify(RG int u, RG int d){
    	Update(rot[u], 0, n - 1, 0, d);
    	for(RG int v = u, dis; frt[v]; v = frt[v]){
    		dis = ChainDiv::Dis(frt[v], u);
    		Update(rot[frt[v]], 0, n - 1, dis, d);
    		Update(rot[v + n], 0, n - 1, dis, d);
    	}
    }
    
    IL int Query(RG int u, RG int k){
    	RG int ans = Sum(rot[u], 0, n - 1, k);
    	for(RG int v = u, dis; frt[v]; v = frt[v]){
    		dis = ChainDiv::Dis(frt[v], u);
    		if(k < dis) continue;
    		ans += Sum(rot[frt[v]], 0, n - 1, k - dis);
    		ans -= Sum(rot[v + n], 0, n - 1, k - dis);
    	}
    	return ans;
    }
    
    int main(RG int argc, RG char* argv[]){
    	sz = n = Read(); Q = Read(); Fill(fst, -1);
    	for(RG int i = 1; i <= n; ++i) w[i] = Read();
    	for(RG int i = 1, a, b; i < n; ++i) a = Read(), b = Read(), Add(a, b), Add(b, a);
    	ChainDiv::Dfs1(1); ChainDiv::Dfs2(1, 1);
    	mx[0] = n + 1; cnt = 0;
    	Getroot(1, 0); root = rt; Create(rt, 0);
    	for(RG int i = 1; i <= n; ++i) Modify(i, w[i]);
    	for(RG int ans = 0; Q; --Q){
    		RG int op = Read(), x = Read() ^ ans, y = Read() ^ ans;
    		if(!op) printf("%d
    ", ans = Query(x, y));
    		else Modify(x, y - w[x]), w[x] = y;
    	}
        return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/cjoieryl/p/8279257.html
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