Description
N个点,形成一个树状结构。有M次发放,每次选择两个点x,y
对于x到y的路径上(含x,y)每个点发一袋Z类型的物品。完成
所有发放后,每个点存放最多的是哪种物品。
Input
第一行数字N,M
接下来N-1行,每行两个数字a,b,表示a与b间有一条边
再接下来M行,每行三个数字x,y,z.如题
Output
输出有N行
每i行的数字表示第i个点存放最多的物品是哪一种,如果有
多种物品的数量一样,输出编号最小的。如果某个点没有物品
则输出0
题解
把操作差分一下,然后就可以用线段树合并维护了。
模板题
线段树合并用来维护与子树中物品个数、子树大小有关的信息
#include<bits/stdc++.h>
using namespace std;
#define rep(i,l,r) for(register int i = l ; i <= r ; i++)
#define repd(i,r,l) for(register int i = r ; i >= l ; i--)
#define rvc(i,S) for(register int i = 0 ; i < (int)S.size() ; i++)
#define rvcd(i,S) for(register int i = ((int)S.size()) - 1 ; i >= 0 ; i--)
#define fore(i,x)for (register int i = head[x] ; i ; i = e[i].next)
#define forup(i,l,r) for (register int i = l ; i <= r ; i += lowbit(i))
#define fordown(i,id) for (register int i = id ; i ; i -= lowbit(i))
#define pb push_back
#define prev prev_
#define stack stack_
#define mp make_pair
#define fi first
#define se second
#define lowbit(x) (x&(-x))
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef unsigned int ui;
typedef pair<int,int> pr;
const int maxn = 100020;
struct node{
int next,to;
}e[maxn * 2];
int head[maxn],cnt,dth[maxn],jump[20][maxn],fa[maxn],rt[maxn],n,m,a[maxn],ans[maxn];
vector <int> add[maxn],del[maxn];
namespace MergeSeg{
//#define ls(x) (x << 1)
//#define rs(x) ((x << 1) | 1)
//每次合并和修改新建节点,空间nlogn
const int N = 1e5 + 10;
int mx[N * 30],id[N * 30],ls[N * 30],rs[N * 30],tot;
inline void update(int x){
if ( mx[ls[x]] >= mx[rs[x]] ) id[x] = id[ls[x]] , mx[x] = mx[ls[x]];
else id[x] = id[rs[x]] , mx[x] = mx[rs[x]];
}
inline void copy(int c,int x){
mx[c] = mx[x] , id[c] = id[x] , ls[c] = ls[x] , rs[c] = rs[x];
}
void modify(int &x,int l,int r,int p,int d){
// copy(++tot,x) , x = tot;
if ( !x ) x = ++tot;
if ( l == r ){
mx[x] += d;
id[x] = l;
return;
}
int mid = (l + r) >> 1;
if ( p <= mid ) modify(ls[x],l,mid,p,d);
else modify(rs[x],mid + 1,r,p,d);
update(x);
}
/* int merge(int x,int y,int l,int r){
if ( !x && !y ) return 0;
int c = ++tot;
if ( !x ){ copy(c,y); return c; }
if ( !y ){ copy(c,x); return c; }
if ( l == r ){
mx[c] = mx[x] + mx[y];
return c;
}
cur1 += (ll)mx[rs[x]] * mx[ls[y]];
cur2 += (ll)mx[ls[x]] * mx[rs[y]];
int mid = (l + r) >> 1;
ls[c] = merge(ls[x],ls[y],l,mid);
rs[c] = merge(rs[x],rs[y],mid + 1,r);
update(c);
return c;
}*/
int merge(int x,int y,int l,int r){ //如果不需要保存子树信息用于将来查询则可以不新建节点
if ( !x && !y ) return 0;
if ( !x ){ return y; }
if ( !y ){ return x; }
if ( l == r ){
mx[x] = mx[x] + mx[y];
id[x] = l;
return x;
}
int mid = (l + r) >> 1;
ls[x] = merge(ls[x],ls[y],l,mid);
rs[x] = merge(rs[x],rs[y],mid + 1,r);
update(x);
return x;
}
}
using namespace MergeSeg;
inline void adde(int x,int y){
e[++cnt].to = y;
e[cnt].next = head[x];
head[x] = cnt;
}
inline int lca(int x,int y){
if ( dth[x] < dth[y] ) swap(x,y);
int d = dth[x] - dth[y];
rep(i,0,19) if ( d & (1 << i) ) x = jump[i][x];
if ( x == y ) return x;
repd(i,19,0) if ( jump[i][x] != jump[i][y] ) x = jump[i][x] , y = jump[i][y];
return fa[x];
}
void dfs(int x){
fore(i,x){
if ( e[i].to == fa[x] ) continue;
jump[0][e[i].to] = fa[e[i].to] = x , dth[e[i].to] = dth[x] + 1;
dfs(e[i].to);
}
}
inline int get(int x){
return lower_bound(a + 1,a + m + 1,x) - a;
}
void dfs_ans(int x){
fore(i,x){
if ( e[i].to == fa[x] ) continue;
dfs_ans(e[i].to);
rt[x] = merge(rt[x],rt[e[i].to],1,m);
}
rvc(i,add[x]){
int tp = get(add[x][i]);
modify(rt[x],1,m,tp,1);
}
rvc(i,del[x]){
int tp = get(del[x][i]);
modify(rt[x],1,m,tp,-1);
}
ans[x] = a[id[rt[x]]];
}
int main(){
scanf("%d %d",&n,&m);
rep(i,1,n - 1){
int x,y;
scanf("%d %d",&x,&y);
adde(x,y),adde(y,x);
}
dfs(1);
rep(i,1,19) rep(j,1,n) jump[i][j] = jump[i - 1][jump[i - 1][j]];
rep(i,1,m){
int x,y,z;
scanf("%d %d %d",&x,&y,&z);
int lca_ = lca(x,y);
add[x].pb(z) , add[y].pb(z);
del[lca_].pb(z) , del[fa[lca_]].pb(z);
a[i] = z;
}
sort(a + 1,a + m + 1);
dfs_ans(1);
rep(i,1,n) printf("%d\n",ans[i]);
}