[HDOJ 4918] Query on the subtree [树的分治+树状数组]

给一颗树，每个点上有权值。有两种操作，修改某个点的权值，或者查询距离某个点不超过d的点的权值和。

数据范围：树的规模和操作次数均为10^5。

树的分治，对于每个分治中心以及他们的孩子，维护树状数组，距离他们d的点的权值和。

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <algorithm>#include <vector>#include <iostream>using namespace std;struct BIT {vector<int> a;void clear(int n) {a.clear();a.resize(n+1,0);}int lb(int x) {return x&-x;}int get(int i) {if (i>=a.size()) i=a.size()-1;int ans=0;for (;i>0;i-=lb(i)) ans+=a[i];return ans;}void set(int i,int x) {for (;i<a.size();i+=lb(i)) a[i]+=x;}void add(BIT &b) {int last=0;for (int i=1;i<b.a.size();i++) {int tmp=b.get(i);set(i,tmp-last);last=tmp;}}};struct Node {int belong[20],from[20],dis[20];BIT num[20];int v,fe,n;bool visited;void clear() {fe=-1;visited=false;}};struct Edge {int t,ne;};Node a[100001];Edge b[200000];int bp;void putedge(int x,int y) {b[bp].t=y;b[bp].ne=a[x].fe;a[x].fe=bp++;}int center,centerV;void getCenter(int i,int n) {a[i].visited=true;int maxn=0;a[i].n=1;for (int j=a[i].fe;j!=-1;j=b[j].ne) {if (!a[b[j].t].visited) {getCenter(b[j].t,n);a[i].n+=a[b[j].t].n;maxn=max(maxn,a[b[j].t].n);}}maxn=max(maxn,n-a[i].n);if (maxn<centerV) {centerV=maxn;center=i;}a[i].visited=false;}void calNum(int i,int kk,int from,int belong,int dis) {a[i].visited=true;a[i].dis[kk]=dis;a[i].belong[kk]=belong;a[i].from[kk]=from;a[from].num[kk].set(dis,a[i].v);for (int j=a[i].fe;j!=-1;j=b[j].ne) {if (!a[b[j].t].visited) {calNum(b[j].t,kk,from,belong,dis+1);}}a[i].visited=false;}void calAns(int i,int n,int kk) {centerV=10000000;getCenter(i,n);i=center;a[i].num[kk].clear(centerV+1);getCenter(i,n);a[i].visited=true;a[i].belong[kk]=i;for (int j=a[i].fe;j!=-1;j=b[j].ne) {if (!a[b[j].t].visited) {a[b[j].t].num[kk].clear(a[b[j].t].n+1);calNum(b[j].t,kk,b[j].t,i,1);a[i].num[kk].add(a[b[j].t].num[kk]);}}for (int j=a[i].fe;j!=-1;j=b[j].ne) if (!a[b[j].t].visited) calAns(b[j].t,a[b[j].t].n,kk+1);}void change(int x,int y) {int k=0;while (a[x].belong[k]!=x) {a[a[x].belong[k]].num[k].set(a[x].dis[k],y-a[x].v);a[a[x].from[k]].num[k].set(a[x].dis[k],y-a[x].v);k++;}a[x].v=y;}int get(int x,int y) {int k=0,ans=0;while (a[x].belong[k]!=x) {int dis=a[x].dis[k];int i=a[x].belong[k];int j=a[x].from[k];if (y>dis) {ans+=a[i].num[k].get(y-dis);ans-=a[j].num[k].get(y-dis);}if (y>=dis) ans+=a[i].v;k++;}ans+=a[x].num[k].get(y);ans+=a[x].v;return ans;}void print(int n) {int i,j,k;printf("PRINT\n");for (i=1;i<=n;i++) {printf("Node %d:\n",i);printf(" belong:");for (j=0;j<3;j++) printf(" %d",a[i].belong[j]);printf("\n from:");for (j=0;j<3;j++) printf(" %d",a[i].from[j]);printf("\n dis:");for (j=0;j<3;j++) printf(" %d",a[i].dis[j]);printf("\n num:");for (j=0;j<3;j++) {printf("\n  %d:",j);for (k=1;k<a[i].num[j].a.size();k++)printf(" %d",a[i].num[j].get(k)-a[i].num[j].get(k-1));}printf("\n");}}int main() {int n,q,i,cas=1;while (scanf("%d%d",&n,&q)!=EOF) {//cerr << "Case: " << cas++ << endl;for (i=1;i<=n;i++) {a[i].clear();scanf("%d",&a[i].v);}bp=0;for (i=1;i<n;i++) {int x,y;scanf("%d%d",&x,&y);putedge(x,y);putedge(y,x);}calAns(1,n,0);for (i=0;i<q;i++) {char c;int x,y;scanf(" %c%d%d",&c,&x,&y);//cerr << cas << ' ' << i << endl;//cerr << c << ' ' << x << ' ' << y << endl;if (c=='!') change(x,y);else printf("%d\n",get(x,y));//print(n);}}return 0;}