UVA 1674 Lightning Energy Report (树链剖分)

题意：

给你一颗树（节点数最多5w）， 给你q个操作，  每个操作，让你u，v两个结点之间的路径上的所有的点的权值都加上w。最后输出n 个点的权值？

思路：

裸树链剖分= =~

和HDU 3966 一样的， 用树状数组就可以了

详细见那一篇博客：Blog_HDU 3966

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#define ps push_back
#define Siz(x) (int)x.size()
using namespace std;

const int maxn = 50000 + 7;

int T, n, q, id, ks;
int fa[maxn], son[maxn], num[maxn], p[maxn], fp[maxn], top[maxn], c[maxn], deep[maxn];

vector<int>g[maxn];

int lowbit(int x){
    return x&-x;
}
int sum(int i){
    int ans = 0;
    while(i){
        ans += c[i];
        i -= lowbit(i);
    }
    return ans;
}

void add(int i,int val){
    while(i <= n){
        c[i] += val;
        i += lowbit(i);
    }
}
void init(){
    id = 1;
    for (int i = 0; i < n; ++i)g[i].clear();
    memset(son,-1,sizeof son);
    memset(c,0,sizeof c);


}
void dfs(int cur,int f,int dep){
    fa[cur] = f;
    num[cur] = 1;
    deep[cur] = dep;
    for (int i = 0; i < Siz(g[cur]); ++i){
        int v = g[cur][i];
        if (v != f){
            dfs(v,cur,dep+1);
            if (son[cur] == -1 || num[v] > num[cur]){
                son[cur] = v;
                num[cur] = 1 + num[v];
            }
        }
    }
}

void link(int cur,int tp){
    p[cur] = id++;
    fp[id-1] = cur;
    top[cur] = tp;
    if (son[cur] == -1) return;
    link(son[cur],tp);
    for (int i = 0; i < Siz(g[cur]); ++i){
        int v = g[cur][i];
        if (v != fa[cur] && v != son[cur]) link(v,v);
    }

}

void deal(int u,int v,int w){
    int fu = top[u], fv = top[v];
    while(fu != fv){
        if (deep[fu] < deep[fv]){
            swap(fu,fv);
            swap(u,v);
        }
        add(p[fu],w); add(p[u]+1,-w);
        u = fa[fu];
        fu = top[u];
    }
    if (deep[u] < deep[v]) swap(u,v);
    add(p[u]+1,-w);
    add(p[v],w);
}



int main(){

    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        init();
        for (int i = 1; i < n; ++i){
            int u,v;
            scanf("%d %d",&u, &v);
            g[u].ps(v);
            g[v].ps(u);
        }
        dfs(0,0,0);
        link(0,0);
        int q;
        scanf("%d",&q);
        while(q--){
            int u,v,w;
            scanf("%d %d %d",&u, &v, &w);
            deal(u,v,w);
        }
        PRintf("Case #%d:\n",++ks);
        for (int i = 0; i < n; ++i) printf("%d\n",sum(p[i]));

    }
    return 0;
}