hdu3966 Aragorn's Story

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Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4217    Accepted Submission(s): 1164

3 2 5
1 2 3
2 1
2 3
I 1 3 5
Q 2
D 1 2 2
Q 1 
Q 3

7
4
8
Hint
1.The number of enemies may be negative.

2.Huge input, be careful. 
 

题意：给一棵树，并给定各个点权的值，然后有3种操作：

I C1 C2 K: 把C1与C2的路径上的所有点权值加上K

D C1 C2 K：把C1与C2的路径上的所有点权值减去K

Q C：查询节点编号为C的权值

大胆地用C++提交吧！

#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
#include<stack>
#include<set>
using namespace std;
int head[50010],tail;
struct Edge
{
	int to,next;	
}edge[50000*2+10];
void add(int from,int to)
{
	edge[tail].to=to;
	edge[tail].next=head[from];
	head[from]=tail++;
}
int siz[50010],dep[50010],fa[50010],son[50010];
void dfs1(int from)
{
	siz[from]=1;
	for(int i=head[from];i!=-1;i=edge[i].next)
	{
		int to=edge[i].to;
		if(to!=fa[from])
		{
			fa[to]=from;
			dep[to]=dep[from]+1;
			dfs1(to);
			siz[from]+=siz[to];
			if(siz[to]>siz[son[from]])
				son[from]=to;
		}
	}
}
int w[50010],top[50010],cnt;
void dfs2(int from,int tp)
{
	w[from]=++cnt;
	top[from]=tp;
	if(son[from]==0)
		return;
	dfs2(son[from],tp);
	for(int i=head[from];i!=-1;i=edge[i].next)
	{
		int to=edge[i].to;
		if(to!=son[from]&&to!=fa[from])
			dfs2(to,to);
	}
}
struct Tree
{
	int l,r,m,v;
}tree[50000*4+10];
void build(int l,int r,int k)
{
	tree[k].l=l;
	tree[k].r=r;
	tree[k].m=l+r>>1;
	tree[k].v=0;
	if(l==r)
		return;
	build(l,tree[k].m,k<<1);
	build(tree[k].m+1,r,k<<1|1);
}
void update(int l,int r,int val,int k)
{
	if(l==tree[k].l&&r==tree[k].r)
	{
		tree[k].v+=val;
		return;
	}
	if(tree[k].v!=0)
	{
		tree[k<<1].v+=tree[k].v;
		tree[k<<1|1].v+=tree[k].v;
		tree[k].v=0;
	}
	if(r<=tree[k].m)
		update(l,r,val,k<<1);
	else if(l>tree[k].m)
		update(l,r,val,k<<1|1);
	else
	{
		update(l,tree[k].m,val,k<<1);
		update(tree[k].m+1,r,val,k<<1|1);
	}
}
int seek(int x,int k)
{
	if(tree[k].l==tree[k].r&&tree[k].l==x)
		return tree[k].v;
	if(tree[k].v!=0)
	{
		tree[k<<1].v+=tree[k].v;
		tree[k<<1|1].v+=tree[k].v;
		tree[k].v=0;
	}
	if(x<=tree[k].m)
		return seek(x,k<<1);
	return seek(x,k<<1|1);
}
void update(int x,int y,int val)
{
	while(top[x]!=top[y])
	{
		if(dep[top[x]]<dep[top[y]])
			swap(x,y);
		update(w[top[x]],w[x],val,1);
		x=fa[top[x]];
	}
	int l=w[x],r=w[y];
	if(l>r)
		swap(l,r);
	update(l,r,val,1);
}
int num[50010];
char od[2];
int main()
{
	int n,m,p;
	while(scanf("%d%d%d",&n,&m,&p)!=EOF)
	{
		for(int i=1;i<=n;i++)
			scanf("%d",num+i);
		tail=0;
		memset(head,-1,sizeof(head));
		while(m--)
		{
			int from,to;
			scanf("%d%d",&from,&to);
			add(from,to);
			add(to,from);
		}
		memset(son,0,sizeof(son));
		dfs1(1);
		cnt=0;
		dfs2(1,1);
		build(1,cnt,1);
		for(int i=1;i<=n;i++)
			update(i,i,num[i]);
		while(p--)
		{
			scanf("%s",od);
			if(od[0]=='Q')
			{
				int x;
				scanf("%d",&x);
				printf("%d\n",seek(w[x],1));
			}
			else
			{
				int l,r,v;
				scanf("%d%d%d",&l,&r,&v);
				if(od[0]=='D')
					v=-v;
				update(l,r,v);
			}
		}
	}
}

hdu3966 Aragorn's Story

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原文地址：http://blog.csdn.net/stl112514/article/details/44758921