清明梦超能力者黄YY

题目连接

https://www.nowcoder.com/acm/contest/206/I

暂时有两种做法.

算法一

涉及:树链剖分,扫描线

在一个线段的情况下,我们可以把一个染色区间拆成左端点处增加事件,右端点处删除事件.
维护一颗权值线段树.
这样,端点从小到大扫描时,遇到增加事件就在线段树指定位置+1,遇到删除事件就在线段树指定位置-1.

那么要回答一个点的答案只需要扫到这个点的时候,在权值线段树里二分$k$大值即可.

这个搬到了树上,我们照样可以使用扫描线的方法来做.

对于每一个染色区间,将其剖到树链上去,可以剖出最多$log(n)$个小区间.

这些小区间的深度较小的端点加入增加事件,深度较大的端点加入删除事件.

然后对整棵树按照$dfn$序列从小到大依次扫描即可,注意扫描的过程中也要维护一颗权值线段树.

算法二

涉及:dfs,权值线段树合并

对于每个询问$u,v$,应该在$u$处加入增加事件,在$v$处加入增加事件,在$lca(u,v)$处加入删除事件,在$faz[lca(u,v)]$处加入删除事件.因此每个询问会产生$4$个事件.

我们采用dfs的顺序,每个点维护一颗权值线段树,然后后序遍历到达一个节点时候,先将它所有儿子的线段树进行合并,合并完成后,将该点涉及到的增加或删除事件执行到线段树上,此时,该点的权值线段树维护的就是横跨这个节点的所有染色区间.

然后在权值线段树上进行$k$大询问即可得到该点的答案.

算法一.代码

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#define pr(x) std::cout << #x << ':' << x << std::endl
#define rep(i,a,b) for(int i = a;i <= b;++i)
#define clr(x) memset(x,0,sizeof(x))const int N = 100007;struct edge{int v,nxt;
}es[N<<1];int head[N],tot;void addedge(int u,int v){es[tot].v = v;es[tot].nxt = head[u];head[u] = tot++;
}int faz[N],siz[N],dep[N],top[N],son[N],dfn[N],rnk[N],idx;void dfs1(int u,int fa,int deep) {faz[u] = fa;siz[u] = 1;dep[u] = deep;for(int e = head[u];e != -1;e = es[e].nxt) {int v = es[e].v;if(v == fa) continue;dfs1(v,u,deep+1);siz[u] += siz[v];if(!son[u] || siz[v] > siz[son[u]])son[u] = v;}
}void dfs2(int u,int tp) {dfn[u] = ++idx;rnk[idx] = u;top[u] = tp;if(son[u]) dfs2(son[u],tp);for(int e = head[u];e != -1;e = es[e].nxt) {int v = es[e].v;if(son[u] != v && faz[u] != v)dfs2(v,v);}
}struct node{int val,len;node(int val = 0,int len = 1):val(val),len(len){}
}ns[N<<2];node maintain(node &lch,node &rch) {return node(lch.val + rch.val,lch.len + rch.len);
}void change(int o,int l,int r,int pos,int val) {if(l == r) {ns[o].val += val;return ;}int mid = (l + r) >> 1;if(pos <= mid)change(o<<1,l,mid,pos,val);elsechange(o<<1|1,mid+1,r,pos,val);ns[o] = maintain(ns[o<<1],ns[o<<1|1]);
}int kth(int o,int l,int r,int x) {if(ns[o].val < x) return 0;if(l == r) return l;   int mid = (l + r) >> 1;if(x > ns[o<<1].val)return kth(o<<1|1,mid+1,r,x-ns[o<<1].val);elsereturn kth(o<<1,l,mid,x);
}struct event{int tp,val;event(int tp = 0,int val = 0):tp(tp),val(val){}
};std::vector<event> events[N];int color[N];void modify(int x,int y,int id) {while(top[x] != top[y]) {if(dep[top[x]] < dep[top[y]])std::swap(x,y);events[dfn[x]].push_back(event(-1,id));events[dfn[top[x]]].push_back(event(1,id));x = faz[top[x]];}if(dep[x] < dep[y])std::swap(x,y);events[dfn[x]].push_back(event(-1,id));events[dfn[y]].push_back(event(1,id));
}int n,m,k;
int ans[N];
int main() {memset(head,-1,sizeof(head));std::ios::sync_with_stdio(false);std::cin >> n >> m >> k;rep(i,1,n-1) {int x,y;std::cin >> x >> y;addedge(x,y);addedge(y,x);}dfs1(1,-1,0);dfs2(1,1);rep(i,1,m){int u,v,c;std::cin >> u >> v >> c;color[m+1-i] = c;modify(u,v,m+1-i);}rep(i,1,n) {for(auto e : events[i]){if(e.tp == 1)change(1,1,n,e.val,1);}ans[rnk[i]] = color[kth(1,1,n,k)];for(auto e : events[i]) {if(e.tp == -1)change(1,1,n,e.val,-1);}}rep(i,1,n) {std::cout << ans[i] << " ";}return 0;
}

算法二.代码

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <assert.h>
#define pr(x) std::cout << #x << ':' << x << std::endl
#define rep(i,a,b) for(int i = a;i <= b;++i)
#define clr(x) memset(x,0,sizeof(x))
#define setinf(x) memset(x,0x3f,sizeof(x))const int N = 100007;
std::vector<int> edge[N];struct node {int sum,lch,rch;node(int sum = 0,int lch = 0,int rch = 0):sum(sum),lch(lch),rch(rch){}
}ns[N*60];
int tot;node maintain(node &lch,node &rch,int l,int r) {return node(lch.sum+rch.sum,l,r);
}void insert(int& o,int l,int r,int pos,int add) {if(!o)o = ++tot;if(l == r) {ns[o].sum += add;return ;}int mid = (l + r) >> 1;if(pos <= mid)insert(ns[o].lch,l,mid,pos,add);elseinsert(ns[o].rch,mid+1,r,pos,add);ns[o] = maintain(ns[ns[o].lch],ns[ns[o].rch],ns[o].lch,ns[o].rch);
}node query(int o,int l,int r,int ql,int qr) {if(o == 0) return ns[0];if(ql <= l && r <= qr)return ns[o];if(r < ql || qr < l)return ns[0];int mid = (l + r) >> 1;node lch = query(ns[o].lch,l,mid,ql,qr);node rch = query(ns[o].rch,mid+1,r,ql,qr);return maintain(lch,rch,0,0);
}int merge(int a,int b) {if(!a) return b;if(!b) return a;int o = ++tot;assert(o < N*60);ns[o].sum = ns[a].sum + ns[b].sum;ns[o].lch = merge(ns[a].lch,ns[b].lch);ns[o].rch = merge(ns[a].rch,ns[b].rch);return o;
}int ans[N],color[N];int faz[N],siz[N],top[N],dep[N],dfn[N],son[N],idx;void dfs1(int u,int fa,int deep) {faz[u] = fa;dep[u] = deep;siz[u] = 1;for(auto v : edge[u]) {if(v == fa) continue;dfs1(v,u,deep+1);siz[u] += siz[v];if(siz[son[u]] < siz[v])son[u] = v;}
}void dfs2(int u,int tp) {dfn[u] = ++idx;top[u] = tp;if(son[u])dfs2(son[u],tp);for(auto v : edge[u]) {if(v == faz[u] || v == son[u]) continue;dfs2(v,v);}
}int lca(int u,int v) {while(top[u] != top[v]) {if(dep[top[u]] < dep[top[v]])v = faz[top[v]];elseu = faz[top[u]];}return dep[u] > dep[v] ? v : u;
}int kth(int o,int l,int r,int k) {if(!o || ns[o].sum < k)return 0;if(l == r) return l;int mid = (l + r) >> 1;if(k <= ns[ns[o].lch].sum)return kth(ns[o].lch,l,mid,k);elsereturn kth(ns[o].rch,mid+1,r,k-ns[ns[o].lch].sum);
}int n,m,k;
int root[N];
std::vector<int> add[N],dec[N];
void dfs(int u,int fa) {for(auto v : edge[u]) {if(v == fa) continue;dfs(v,u);root[u] = merge(root[u],root[v]);}for(auto x : dec[u]){insert(root[u],1,n,x,-1);}for(auto y : add[u]){insert(root[u],1,n,y,1);}ans[u] = color[kth(root[u],1,n,k)];
}int main() {std::ios::sync_with_stdio(false);std::cin >> n >> m >> k;rep(i,1,n-1) {int u,v;std::cin >> u >> v;edge[u].push_back(v);edge[v].push_back(u);}dfs1(1,0,0);dfs2(1,1);for(int i = m;i >= 1;--i) {int u,v;std::cin >> u >> v >> color[i];add[u].push_back(i);add[v].push_back(i);int _lca = lca(u,v);dec[_lca].push_back(i);dec[faz[_lca]].push_back(i);}dfs(1,0);rep(i,1,n) {std::cout << ans[i] << " ";}std::cout << std::endl;return 0;
}