很无奈，模板又一次无奈的打错了。。不过，很快便找到了。。

题意：给一些边，有一些操作，每次操作，都要在这些边上加上1，求每个边的边权。。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
#define lson id << 1
#define rson id << 1|1
const int M = 100008;
int top[M],siz[M],son[M],father[M],ti[M],dep[M];
int e[M][2],idx,tp;
struct {
    int head;
}H[M];
struct{
    int v,next;
}E[M*2];
void init(){
    memset(H,-1,sizeof(H));
    memset(E,-1,sizeof(E));
    tp = 0,idx = 0;
}
void add(int u,int v){
    E[tp].v =v;
    E[tp].next = H[u].head;
    H[u].head = tp++;
}
void dfs_1(int u,int fa){
    son[u] = 0;siz[u] = 1;father[u] = fa;dep[u] = dep[fa] + 1;
    for(int i=H[u].head;i!=-1;i=E[i].next){
        int v = E[i].v;
        if(v == fa)continue;
        dfs_1(v,u);
        siz[u] += siz[v];
        if(siz[v] > siz[son[u]])son[u] = v;
    }
}
void dfs_2(int u,int fa){
    ti[u] = idx++;
    top[u] = fa;
    if(son[u])dfs_2(son[u],fa);
    for(int i=H[u].head;i!=-1;i=E[i].next){
        int v = E[i].v;
        if(v == father[u]|| v == son[u])continue;
        dfs_2(v,v);
    }
}
struct M_tree{
    int mark,l,r;
    int mid(){
        return (l+r) /2;
    }
}node[M*4];
void  build_tree(int id,int l,int r){
    node[id].l = l;node[id].r = r;node[id].mark = 0;

    if(l == r)return;
    int mid = node[id].mid();
    build_tree(lson,l,mid);
    build_tree(rson,mid+1,r);
}
void push_down(int id){
    if(node[id].l == node[id].r)return;
    if(node[id].mark){
        node[lson].mark += node[id].mark ;
        node[rson].mark += node[id].mark ;
        node[id].mark = 0;
    }
}

void nede(int id,int l,int r){

    if(node[id].l == l&&node[id].r == r){

        node[id].mark += 1;
        return;
    }
    push_down(id);
    int mid = node[id].mid();
    if(r <=mid)nede(lson,l,r);
    else if(l > mid)nede(rson,l,r);
    else{
        nede(lson,l,mid);
        nede(rson,mid+1,r);
    }
}
int ans = 0;
int query(int id,int r){

    if( node[id].l == r&&node[id].r == r)return node[id].mark;
    push_down(id);
    int mid= node[id].mid();
    if(r <=mid)return query(lson,r);
    else return query(rson,r);
}
void negat(int u,int v){
    int f1 = top[u];
    int f2 = top[v];

    while(f1 != f2){
        if(dep[f1] < dep[f2]){
            swap(f1,f2);
            swap(u,v);
        }//cout << ti[f1] << "<---->" << ti[u]<<endl;
        nede(1,ti[f1],ti[u]);
        u = father[f1];f1 = top[u];
    }
    if(u == v)return;
    if(dep[u] > dep[v])swap(u,v);
    nede(1,ti[son[u]],ti[v]);
}
int main()
{
    int u,v,n,m;
    while(~scanf("%d",&n)){
            init();
       for(int i=0;i<n-1;i++){
        scanf("%d%d",&e[i][0],&e[i][1]);
        add(e[i][0],e[i][1]);
        add(e[i][1],e[i][0]);

       }
       dfs_1(1,1);
       dfs_2(1,1);
       build_tree(1,0,idx-1);
//cout <<idx<<endl;
       scanf("%d",&m);
       while(m--){
            scanf("%d%d",&u,&v);

            negat(u,v);
       }
       for(int i=0;i<n-1;i++)
       {
           if(dep[e[i][0]] > dep[e[i][1]]){
            swap(e[i][0],e[i][1]);
           }
       }
       for(int i=0;i<n-1;i++){
            printf("%d",query(1,ti[e[i][1]]));
            if(i!=n-2)printf(" ");
       }
       printf("\n");
    }
}

CodeForces 191C 树链剖分 第4遍