P4556 rain tail segment tree merger

The combined use segment tree, each node maintains a tree line weights, index for the relief food categories, the largest number of maintenance intervals of relief grain number (subscript). So each node and the answer is \ (Tre [rot [the X-]] \) .

Then apply differential thinking tree point, for distribution path \ (U, v \) , we \ (U \) +1 on the (v \) \ +1, in \ (lca (u, v) \ ) at -1, the \ (fa (lca) \) at -1, and the bottom made prefix tree, the tree that was merged segment of the current node information up the final count.

Note that may occur when the merger \ (tre [rot [x] ] \) is not \ (0 \) , but \ (sum [rot [x] ] \) is \ (0 \) case this time that number is \ (tre [rot [x] ] \) of relief food to the number \ (0 \) , so in this case should be \ (ans [x] \) is updated to \ (0 \) (according to meaning of the questions).

#include <cstdio>
#include <algorithm>
#define MAXN 100001
#define LOG 19
#define mxr 100000
using namespace std;
inline int read(){
    char ch=getchar();int s=0;
    while(ch<'0'||ch>'9') ch=getchar();
    while(ch>='0'&&ch<='9') s=s*10+(ch^'0'),ch=getchar();
    return s;
}
int head[MAXN],nxt[MAXN*2],vv[MAXN*2],tot;
inline void add_edge(int u, int v){
    vv[++tot]=v;
    nxt[tot]=head[u];
    head[u]=tot;
}
int n,m;
#define MAXM MAXN*20
int tre[MAXM],sum[MAXM],sl[MAXM],sr[MAXM],cnt;
void push_up(int x){
    if(sl[x]==0){
        sum[x]=sum[sr[x]];
        tre[x]=tre[sr[x]];
        return;
    }
    if(sr[x]==0){
        sum[x]=sum[sl[x]];
        tre[x]=tre[sl[x]];
        return;
    }
    if(sum[sl[x]]>=sum[sr[x]]){
        sum[x]=sum[sl[x]];
        tre[x]=tre[sl[x]];
    }else{
        sum[x]=sum[sr[x]];
        tre[x]=tre[sr[x]];
    }
}
void change(int &x, int l, int r, int pos, int val){
    if(x==0) x=++cnt;
    if(l==r){
        sum[x]+=val;
        tre[x]=pos;
        return;
    }
    int mid=(l+r)>>1;
    if(pos<=mid) change(sl[x], l, mid, pos, val);
    else change(sr[x], mid+1, r, pos, val);
    push_up(x);
}
int merge(int a, int b, int l, int r){
    if(a==0||b==0) return a+b;
    if(l==r){
        sum[a]+=sum[b];
        return a;
    }
    int mid=(l+r)>>1;
    sl[a]=merge(sl[a], sl[b], l, mid);
    sr[a]=merge(sr[a], sr[b], mid+1, r);
    push_up(a);
    return a;
}
int f[MAXN][LOG],dep[MAXN];
void dfs(int u, int fa){
    f[u][0]=fa;
    dep[u]=dep[fa]+1;
    for(int i=1;i<LOG;++i)
        f[u][i]=f[f[u][i-1]][i-1];
    for(int i=head[u];i;i=nxt[i]){
        int v=vv[i];
        if(v==fa) continue;
        dfs(v, u);
    }
}
int lca(int a, int b){
    if(dep[a]<dep[b]) swap(a, b);
    for(int i=LOG-1;i>=0;--i)
        if(dep[f[a][i]]>=dep[b])
            a=f[a][i];
    if(a==b) return a;
    for(int i=LOG-1;i>=0;--i)
        if(f[a][i]!=f[b][i])
            a=f[a][i],b=f[b][i];
    return f[a][0];
}
int rot[MAXN],ans[MAXN];
void calc(int u, int fa){
    for(int i=head[u];i;i=nxt[i]){
        int v=vv[i];
        if(v==fa) continue;
        calc(v, u);
        rot[u]=merge(rot[u], rot[v], 1, mxr);
    }
    ans[u]=tre[rot[u]];
    if(sum[rot[u]]==0) ans[u]=0; // 特判没有救济粮的情况
}
int main(){
    n=read(),m=read();
    for(int i=1;i<n;++i){
        int a=read(),b=read();
        add_edge(a, b);
        add_edge(b, a);
    }
    dfs(1, 0);
    for(int i=1;i<=m;++i){
        int a=read(),b=read(),x=read();
        int l=lca(a, b);
        change(rot[a], 1, mxr, x, 1);
        change(rot[b], 1, mxr, x, 1);
        change(rot[l], 1, mxr, x, -1);
        change(rot[f[l][0]], 1, mxr, x, -1);
    }
    calc(1, 0);
    for(int i=1;i<=n;++i) printf("%d\n", ans[i]);
    return 0;
}