[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;
}