树分治经典题+树状数组（hdu4918）

Query on the subtree

Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 536    Accepted Submission(s): 180

Problem Description

bobo has a tree, whose vertices are conveniently labeled by 1,2,…,n. At the very begining, the i-th vertex is assigned with weight w_i.

There are q operations. Each operations are of the following 2 types:

Change the weight of vertex v into x (denoted as "! v x"),
Ask the total weight of vertices whose distance are no more than d away from vertex v (denoted as "? v d").

Note that the distance between vertex u and v is the number of edges on the shortest path between them.

 

Input

The input consists of several tests. For each tests:

The first line contains n,q (1≤n,q≤10⁵). The second line contains n integers w₁,w₂,…,w_n (0≤w_i≤10⁴). Each of the following (n - 1) lines contain 2 integers a_i,b_idenoting an edge between vertices a_i and b_i (1≤a_i,b_i≤n). Each of the following q lines contain the operations (1≤v≤n,0≤x≤10⁴,0≤d≤n).

 

Output

For each tests:

For each queries, a single number denotes the total weight.

 

Sample Input

4 31 1 1 11 22 33 4? 2 1! 1 0? 2 13 31 2 31 21 3? 1 0? 1 1? 1 2

 

Sample Output

32166

 

Author

Xiaoxu Guo (ftiasch)

 

Source

2014 Multi-University Training Contest 5

题意：给定一棵树，有两种操作，1.修改某个节点的值；2.询问与某个点距离不大于d的节点的权值和是多少

思路：树分治+树状数组，树分治的过程中记录下每个节点对应的重心，他是在这个中心的哪个子树中，他距离中心的距离是多少，树状数组记录这个节点能被哪些节点更新到

树分治的关键是每个节点最多被logn个节点更新到

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;const int maxn=100010;const int INF=1000000000;int N,Q;int cnt,id[maxn];int w[maxn];int pool[maxn*40];//树状数组缓冲池int *tail;int tot,head[maxn];int root,Max;int size[maxn],maxv[maxn];struct BIT{    int *tree;    int n;    void init(int tot)    {        n=tot;        tree=tail;        tail=tail+tot;        memset(tree,0,sizeof(int)*n);    }    void add(int x,int val)    {        while(x<n)        {            tree[x]+=val;            x+=~x&x+1;        }    }    int getsum(int x)    {        x=min(x,n-1);        int sum=0;        while(x>=0)        {            sum+=tree[x];            x-=~x&x+1;        }        return sum;    }}bit[maxn<<1];int fa[maxn];int vis[maxn];struct node{    int v,next;}edge[maxn*2];;void init(){    tot=0;    cnt=0;    tail=pool;    memset(head,-1,sizeof(head));    memset(vis,0,sizeof(vis));}void add_edge(int u,int v){    edge[tot].v=v;    edge[tot].next=head[u];    head[u]=tot++;}int dfssize(int u,int fa){    size[u]=1;    maxv[u]=0;    for(int i=head[u];i!=-1;i=edge[i].next)    {        int v=edge[i].v;        if(v==fa||vis[v])continue;        dfssize(v,u);        size[u]+=size[v];        if(size[v]>maxv[u])maxv[u]=size[v];    }    return size[u];}void dfsroot(int r,int u,int f){    if(size[r]-size[u]>maxv[u])        maxv[u]=size[r]-size[u];    if(maxv[u]<Max)Max=maxv[u],root=u;    for(int i=head[u];i!=-1;i=edge[i].next)    {        int v=edge[i].v;        if(v==f||vis[v])continue;        dfsroot(r,v,u);    }}int que[maxn];int dis[maxn];struct Node{    int root,dis,subtree;    Node(){}    Node(int a,int b,int c):root(a),subtree(b),dis(c){}};vector<Node> V[maxn];void solve(int u,int root,int subtree){    int l,r;    que[l=r=1]=u;    dis[u]=1;    fa[u]=0;    while(l<=r)    {        u=que[l];        V[u].push_back(Node(id[root],subtree,dis[u]));        for(int i=head[u];i!=-1;i=edge[i].next)        {            int v=edge[i].v;            if(vis[v]||v==fa[u])continue;            dis[v]=dis[u]+1;            que[++r]=v;            fa[v]=u;        }        l++;    }    bit[subtree].init(r+1);    for(int i=1;i<=r;i++)    {        int v=que[i];        bit[id[root]].add(dis[v],w[v]);        bit[subtree].add(dis[v],w[v]);    }}void dfs(int u){    Max=INF;    int tot=0;    tot=dfssize(u,0);    dfsroot(u,u,0);    vis[root]=1;    id[root]=cnt++;    V[root].push_back(Node(id[root],-1,0));    bit[id[root]].init(tot);    bit[id[root]].add(0,w[root]);    for(int i=head[root];i!=-1;i=edge[i].next)    {        int v=edge[i].v;        if(vis[v])continue;        solve(v,root,cnt);        cnt++;    }    for(int i=head[root];i!=-1;i=edge[i].next)    {        int v=edge[i].v;        if(vis[v])continue;        dfs(v);    }}int main(){    while(scanf("%d%d",&N,&Q)!=EOF)    {        int u,v;        init();        for(int i=0;i<=N;i++)V[i].clear();        for(int i=1;i<=N;i++)scanf("%d",&w[i]);        for(int i=1;i<N;i++)        {            scanf("%d%d",&u,&v);            add_edge(u,v);            add_edge(v,u);        }        dfs(1);        char op[5];        int x,y;        while(Q--)        {            scanf("%s%d%d",op,&x,&y);            if(op[0]=='!')            {                int cha=y-w[x];                int len=V[x].size();                for(int i=0;i<len;i++)                {                    Node tmp=V[x][i];                    bit[tmp.root].add(tmp.dis,cha);                    if(tmp.subtree!=-1)                        bit[tmp.subtree].add(tmp.dis,cha);//?                }                w[x]+=cha;            }            else            {                int ans=0;                int len=V[x].size();                for(int i=0;i<len;i++)                {                    Node tmp=V[x][i];                    ans+=bit[tmp.root].getsum(y-tmp.dis);                    if(tmp.subtree!=-1)                        ans-=bit[tmp.subtree].getsum(y-tmp.dis);                }                printf("%d\n",ans);            }        }    }    return 0;}