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HDU 3966 Aragorn's Story（树链剖分 模板题）

2024-11-23 08:56:02   202人阅读

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3966


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.

题意：

给出每个点的初始值，和给出一些边，连成一颗树；

接着三种操作：

1、Q X：询问X这个点的值；

2、I  X Y Z：在点X到点Y的路径上所有的点，都增加Z；

3、D  X Y Z：在点X到点Y的路径上所有的点，都减少Z；


代码如下：

#include <cstdio>
#include <cstring>
#define MAXN 50017
#define maxn 100017
int N, M, Q;
int siz[MAXN];//siz[v]表示以v为根的子树的节点数
int dep[MAXN];//dep[v]表示v的深度(根深度为1)
int top[MAXN];//top[v]表示v所在的重链的顶端节点
int fa[MAXN];//fa[v]表示v的父亲
int son[MAXN];//son[v]表示与v在同一重链上的v的儿子节点
int w[MAXN];//w[v]表示v与其父亲节点的连边（姑且称为v的父边）在线段树中的位置
int cont;
int first[maxn], e, next[maxn], v[maxn];
int a[MAXN], add[MAXN*4], q[MAXN];
void swap_(int &x, int &y)
{
    int t;
    t = x, x = y, y = t;
}

void Add(int x, int y)
{
    v[e] = y;
    next[e] = first[x], first[x] = e++;
}
void pushdown(int cur)
{
    if(add[cur])
    {
        add[cur << 1] += add[cur];
        add[cur << 1 | 1] += add[cur];
        add[cur] = 0;
    }
}

void prepare()//剖分部分
{
    int x, rear = 0;
    q[rear++] = 1;
    fa[1] = 0;
    dep[1] = 1;//根深度为1
    memset(top,0,sizeof(top));
    for(int i = 0; i < rear; i++)
    {
        x = q[i];
        for(int j = first[x]; j != -1; j = next[j])
        {
            if(v[j] != fa[x])
            {
                fa[v[j]] = x, dep[v[j]] = dep[x] + 1;
                q[rear++] = v[j];
            }
        }
    }
    siz[0] = 0;
    for(int i = rear-1; i >= 0; i--)
    {
        x = q[i];
        siz[x] = 1, son[x] = 0;
        for(int j = first[x]; j != -1; j = next[j])
        {
            if(v[j] != fa[x])
            {
                siz[x] += siz[v[j]];
                if(siz[v[j]] > siz[son[x]])
                    son[x] = v[j];
            }
        }
    }
    cont = 0;
    for(int i = 0; i < rear; i++)
    {
        x = q[i];
        if(top[x] == 0)
        {
            for(int j = x; j != 0; j = son[j])
            {
                top[j] = x, w[j] = ++cont;
            }
        }
    }
}

void  Update(int cur, int x, int y, int s, int t, int v)//线段树部分
{
    int mid = (x+y) >> 1;
    int ls = cur << 1;
    int rs = cur << 1 | 1;
    if(x >= s && y <= t)
    {
        add[cur] += v;
        return ;
    }
    pushdown(cur);
    if(mid >= s)
        Update(ls, x, mid, s, t, v);
    if(mid < t)
        Update(rs, mid+1, y, s, t, v);
}

int query(int cur, int x, int y, int k)
{
    int mid = (x+y) >> 1;
    int ls = cur << 1;
    int rs = cur << 1 | 1;
    if(x == y)
        return add[cur];
    pushdown(cur);
    if( k <= mid)
        return query(ls, x, mid, k);
    else
        return query(rs, mid+1, y, k);
}

void Deal(int x, int y, int z)
{
    int fx = top[x], fy = top[y];
    while(fx != fy)
    {
        if(dep[fx] > dep[fy])
        {
            swap_(fx, fy);
            swap_(x, y);
        }
        Update(1, 1, N, w[fy], w[y], z);
        y = fa[fy], fy = top[y];
    }
    if(dep[x] > dep[y])
    {
        swap_(x, y);
    }
    Update(1, 1, N, w[x], w[y], z);
}

int main()
{
    int x, y, z;
    while(scanf("%d%d%d",&N,&M,&Q) != EOF)
    {
        for(int i = 1; i <= N; i++)
        {
            scanf("%d",&a[i]);
        }
        e = 0;
        memset(first,-1,sizeof(first));
        for(int i = 0; i < M; i++)
        {
            scanf("%d%d",&x,&y);
            Add(x, y);
            Add(y, x);
        }
        prepare();
        
        memset(add,0,sizeof(add));
        for(int i = 1; i <= N; i++)
        {
            Update(1, 1, N, w[i], w[i], a[i]);
        }
        char op[7];
        for(int i = 1; i <= Q; i++)
        {
            scanf("%s",op);
            if(op[0] == 'Q')
            {
                scanf("%d",&x);
                printf("%d\n",query(1, 1, N, w[x]));
            }
            else
            {
                scanf("%d%d%d",&x,&y, &z);
                if(op[0] == 'I')
                    Deal(x, y, z);
                else
                    Deal(x, y, -z);
            }
        }
    }
    return 0;
}


HDU 3966 Aragorn's Story（树链剖分 模板题）


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