题意：

给定n个点的树，m个操作

树有点权和边权

下面n-1行给出树边

下面m行操作 ：

● ADD1 u v k: for nodes on the path from u to v, the value of these nodes increase by k.

● ADD2 u v k: for edges on the path from u to v, the value of these edges increase by k.
跑个LCA，然后sum0[i]表示i 的点权，lazy0[i]表示 从[i,root] 这条路径上增加的值。

类似于前缀和的思想，最后dfs求个前缀和。

G++栈浅 && 开栈外挂无效，么么哒。。

#pragma comment(linker, "/STACK:1024000000,1024000000")    
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <iostream>
#include <algorithm>
using namespace std;
typedef __int64 ll;
inline void rd(int &n){    
    n = 0;
    char c = getchar();    
    while(c < '0' || c > '9') c = getchar();       
    while(c >= '0' && c <= '9') n *= 10, n += (c - '0'),c = getchar();    
}    
inline void rd64(ll &n){    
    ll x = 0, tmp = 1;    
    char c = getchar();    
    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();      
    if(c == '-') c = getchar(), tmp = -1;    
    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();    
    n = x*tmp;   
}  
int data[30];
inline void pt(ll n){    
    if(n < 0)  
        putchar('-'), n = -n;    
    int len = 0;    
    while(n)    
        data[len++] = n%10, n /= 10;     
    if(!len) data[len++] = 0;    
    while(len--) putchar(data[len]+48);    
}  
#define N 100005
struct Edge{    
    int to, nex;    
}edge[2*N];    
int head[N],edgenum,fa[N][20],dep[N];  //fa[i][x] 是i的第2^x个父亲（如果超过范围就是根）  
void add(int u,int v){    
    Edge E={v,head[u]};  
    edge[edgenum] = E;  
    head[u]=edgenum++;    
}  
void bfs(int root){    
    queue<int> q;    
    fa[root][0]=root;dep[root]=0;   
    q.push(root);    
    while(!q.empty()){    
        int u=q.front();q.pop();    
        for(int i=1;i<20;i++)fa[u][i]=fa[fa[u][i-1]][i-1];    
        for(int i=head[u]; ~i;i=edge[i].nex){    
            int v=edge[i].to;if(v==fa[u][0])continue;    
            dep[v]=dep[u]+1; fa[v][0]=u;    
            q.push(v);  
        }    
    }    
}    
int Lca(int x,int y){    
    if(dep[x]<dep[y])swap(x,y);    
    for(int i=0;i<20;i++)if((dep[x]-dep[y])&(1<<i))x=fa[x][i];    
    if(x==y)return x;    
    for(int i=19;i>=0;i--)if(fa[x][i]!=fa[y][i])x=fa[x][i],y=fa[y][i];    
    return fa[x][0];  
}    
void init(){memset(head, -1, sizeof head); edgenum = 0;}  
int n, que;
ll sum0[N], lazy0[N], sum1[N], lazy1[N];
void dfs(int u, int Father){
    sum0[u] += lazy0[u];
    sum1[u] += lazy1[u];
    for(int i = head[u]; ~i; i = edge[i].nex) {
        int v = edge[i].to; if( v == Father ) continue;
        dfs(v, u);
        sum0[u] += lazy0[v];
        sum1[u] += lazy1[v];
        lazy0[u] += lazy0[v];
        lazy1[u] += lazy1[v];
    }
}
void input() {
    memset(sum0, 0, sizeof sum0);
    memset(sum1, 0, sizeof sum1);
    memset(lazy0, 0, sizeof lazy0);
    memset(lazy1, 0, sizeof lazy1);
    rd(n); rd(que);
    init();
    for(int i = 1; i < n; i++) {
        int u, v;
        rd(u); rd(v);
        add(u, v); add(v, u);
    }
    bfs(1);
}
int main() {
    int T, Cas = 1; rd(T);
    while (T -- )
    {
        input();
        printf("Case #%d:\n", Cas++);
        char op[6]; int l, r; ll val;
        while(que--){
            scanf("%s", op); rd(l); rd(r); rd64(val);
            if(dep[l] < dep[r]) swap(l, r); //让l在下面
            if(op[3] == '1') 
            {
                int LCA = Lca(l, r);
                if(LCA == r) {
                    lazy0[l] += val; //从[l, 1]上增加val
                    lazy0[r] -= val;
                    sum0[r] += val;
                }
                else {
                    lazy0[l] += val;
                    lazy0[r] += val;
                    lazy0[LCA] -= val*2LL;
                    sum0[LCA] += val;
                }
            }
            else
            {
                int LCA = Lca(l, r);
                if(LCA == r) {
                    lazy1[l] += val; //从[l, 1]上增加val
                    lazy1[r] -= val;
                }
                else {
                    lazy1[l] += val;
                    lazy1[r] += val;
                    lazy1[LCA] -= val*2LL;
                }
            }
        }
        dfs(1, 1);
        for(int i = 1; i <= n; i++) {
            pt(sum0[i]);
            putchar(i==n?'\n':' ');
        }
        int fir = 0;
        for(int i = 0; i < edgenum; i+=2) {
            int u = edge[i].to, v = edge[i^1].to;
            if(dep[u] < dep[v])
                u = v;
            if(fir++)putchar(' ');
            pt(sum1[u]);
        }
        puts("");
    }
    return 0;
}
/*
99
9 4
1 2
2 3
3 5
5 6
5 7
2 4
4 8
4 9
ADD1 1 9 10
ADD1 1 7 5
ADD1 6 9 3
ADD1 8 7 100

5 4
1 2 
2 3
2 4
2 5
ADD1 1 4 1
ADD1 5 3 1
ADD2 5 2 2
ADD2 2 4 1



*/

HDU Tree LCA 2014 ACM/ICPC Asia Regional Shanghai Online