HDU 3966 Aragorn's Story （树链剖分+树状数组）

Aragorn's Story

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1321    Accepted Submission(s): 344

Problem Description

Our protagonist is the handsome human prince Aragorn comes from The Lord of the Rings. One day Aragorn finds a lot of enemies who want to invade his kingdom. As Aragorn knows, the enemy has N camps out of his kingdom and M edges connect them. It is guaranteed that for any two camps, there is one and only one path connect them. At first Aragorn know the number of enemies in every camp. But the enemy is cunning , they will increase or decrease the number of soldiers in camps. Every time the enemy change the number of soldiers, they will set two camps C1 and C2. Then, for C1, C2 and all camps on the path from C1 to C2, they will increase or decrease K soldiers to these camps. Now Aragorn wants to know the number of soldiers in some particular camps real-time.

 

 

Input

Multiple test cases, process to the end of input.

For each case, The first line contains three integers N, M, P which means there will be N(1 ≤ N ≤ 50000) camps, M(M = N-1) edges and P(1 ≤ P ≤ 100000) operations. The number of camps starts from 1.

The next line contains N integers A1, A2, ...AN(0 ≤ Ai ≤ 1000), means at first in camp-i has Ai enemies.

The next M lines contains two integers u and v for each, denotes that there is an edge connects camp-u and camp-v.

The next P lines will start with a capital letter 'I', 'D' or 'Q' for each line.

'I', followed by three integers C1, C2 and K( 0≤K≤1000), which means for camp C1, C2 and all camps on the path from C1 to C2, increase K soldiers to these camps.

'D', followed by three integers C1, C2 and K( 0≤K≤1000), which means for camp C1, C2 and all camps on the path from C1 to C2, decrease K soldiers to these camps.

'Q', followed by one integer C, which is a query and means Aragorn wants to know the number of enemies in camp C at that time.

 

 

Output

For each query, you need to output the actually number of enemies in the specified camp.

 

 

Sample Input

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

 

 

Sample Output

7 4 8

Hint

1.The number of enemies may be negative. 2.Huge input, be careful.

 

 

Source

2011 Multi-University Training Contest 13 - Host by HIT

 

 

 

 

裸的树链剖分的入门题；

我是套的树状数组实现的

 

/* ***********************************************
Author        :kuangbin
Created Time  :2013/8/14 23:14:27
File Name     :F:\2013ACM练习\专题学习\数链剖分\HDU3966.cpp
************************************************ */
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int MAXN = 50010;
struct Edge
{
    int to,next;
}edge[MAXN*2];
int head[MAXN],tot;
int top[MAXN];
int fa[MAXN];
int deep[MAXN];
int num[MAXN];
int p[MAXN];
int fp[MAXN];
int son[MAXN];
int pos;
void init()
{
    tot = 0;
    memset(head,-1,sizeof(head));
    pos = 1;//使用树状数组，编号从头1开始
    memset(son,-1,sizeof(son));
}
void addedge(int u,int v)
{
    edge[tot].to = v; edge[tot].next = head[u]; head[u] = tot++;
}
void dfs1(int u,int pre,int d)
{
    deep[u] = d;
    fa[u] = pre;
    num[u] = 1;
    for(int i = head[u];i != -1; i = edge[i].next)
    {
        int v = edge[i].to;
        if(v != pre)
        {
            dfs1(v,u,d+1);
            num[u] += num[v];
            if(son[u] == -1 || num[v] > num[son[u]])
                son[u] = v;
        }
    }
}
void getpos(int u,int sp)
{
    top[u] = sp;
    p[u] = pos++;
    fp[p[u]] = u;
    if(son[u] == -1) return;
    getpos(son[u],sp);
    for(int i = head[u];i != -1;i = edge[i].next)
    {
        int v = edge[i].to;
        if( v != son[u] && v != fa[u])
            getpos(v,v);
    }
}

//树状数组
int lowbit(int x)
{
    return x&(-x);
}
int c[MAXN];
int n;
int sum(int i)
{
    int s = 0;
    while(i > 0)
    {
        s += c[i];
        i -= lowbit(i);
    }
    return s;
}
void add(int i,int val)
{
    while(i <= n)
    {
        c[i] += val;
        i += lowbit(i);
    }
}
void Change(int u,int v,int val)//u->v的路径上点的值改变val
{
    int f1 = top[u], f2 = top[v];
    int tmp = 0;
    while(f1 != f2)
    {
        if(deep[f1] < deep[f2])
        {
            swap(f1,f2);
            swap(u,v);
        }
        add(p[f1],val);
        add(p[u]+1,-val);
        u = fa[f1];
        f1 = top[u];
    }
    if(deep[u] > deep[v]) swap(u,v);
    add(p[u],val);
    add(p[v]+1,-val);
}
int a[MAXN];
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
     int M,P;
    while(scanf("%d%d%d",&n,&M,&P) == 3)
    {
        int u,v;
        int C1,C2,K;
        char op[10];
        init();
        for(int i = 1;i <= n;i++)
        {
            scanf("%d",&a[i]);
        }
        while(M--)
        {
            scanf("%d%d",&u,&v);
            addedge(u,v);
            addedge(v,u);
        }
        dfs1(1,0,0);
        getpos(1,1);
        memset(c,0,sizeof(c));
        for(int i = 1;i <= n;i++)
        {
            add(p[i],a[i]);
            add(p[i]+1,-a[i]);
        }
        while(P--)
        {
            scanf("%s",op);
            if(op[0] == 'Q')
            {
                scanf("%d",&u);
                printf("%d\n",sum(p[u]));
            }
            else
            {
                scanf("%d%d%d",&C1,&C2,&K);
                if(op[0] == 'D')
                    K = -K;
                Change(C1,C2,K);
            }
        }
    }
    return 0;
}