树链剖分+离散+扫描（HDU5044）

Tree

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1385    Accepted Submission(s): 237

Problem Description

You are given a tree (an acyclic undirected connected graph) with N nodes. The tree nodes are numbered from 1 to N

There are N - 1 edges numbered from 1 to N - 1.

Each node has a value and each edge has a value. The initial value is 0.

There are two kind of operation as follows:

● ADD1 u v k: for nodes on the path from u to v, the value of these nodes increase by k.

● ADD2 u v k: for edges on the path from u to v, the value of these edges increase by k.

After finished M operation on the tree, please output the value of each node and edge.

 

Input

The first line of the input is T (1 ≤ T ≤ 20), which stands for the number of test cases you need to solve.

The first line of each case contains two integers N ,M (1 ≤ N, M ≤10 ⁵),denoting the number of nodes and operations, respectively.

The next N - 1 lines, each lines contains two integers u, v(1 ≤ u, v ≤ N ), denote there is an edge between u,v and its initial value is 0.

For the next M line, contain instructions “ADD1 u v k” or “ADD2 u v k”. (1 ≤ u, v ≤ N, -10 ⁵ ≤ k ≤ 10 ⁵)

 

Output

For each test case, print a line “Case #t:”(without quotes, t means the index of the test case) at the beginning.

The second line contains N integer which means the value of each node.

The third line contains N - 1 integer which means the value of each edge according to the input order.

 

Sample Input

  
    

     
   2
4 2
1 2
2 3
2 4
ADD1 1 4 1
ADD2 3 4 2
4 2
1 2
2 3
1 4
ADD1 1 4 5
ADD2 3 2 4
  
    

 

Sample Output

  
    

     
   Case #1:
1 1 0 1
0 2 2
Case #2:
5 0 0 5
0 4 0
  
    
  
    

   
     题意：给出一棵树，树上的点和边的权值开始都是0，有两种操作，对于第一种操作，ADD1：在u到v的路径上每个点的权值+w；对于第二种操作ADD2：在u到v的路径上每个边的权值+w；最后询问每个点的权值和每条边的权值（按照输入的顺序）
  
    
  
    

   
     分析：首先两个dfs进行轻重链剖分，然后对于每个区间[L,R],在L的位置+w，在R+1的位置-w，保存在g1数组中，同理，对于边也一样，保存在g2数组里，最后从前往后做一遍扫描即可；
  
    
  
    

   
     程序：
  
    
  
    

   
     
   
     
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include"stdio.h"
#include"string.h"
#include"iostream"
#include"map"
#include"string"
#include"queue"
#include"stdlib.h"
#include"algorithm"
#include"math.h"
#define M 110009
#define eps 1e-5
#define inf 100000000
#define mod 100000000
#define INF 0x3f3f3f3f
using namespace std;
struct node
{
    int v;
    node(int vv){v=vv;}
};
vector<node>edge[M];
int pos,son[M],fa[M],p[M],fp[M],deep[M],top[M],num[M],g1[M],g2[M],ans[M];
void dfs(int u,int f,int d)
{
    deep[u]=d;
    fa[u]=f;
    num[1]=1;
    for(int i=0;i<(int)edge[u].size();i++)
    {
        int v=edge[u][i].v;
        if(v==f)continue;
        dfs(v,u,d+1);
        num[u]+=num[v];
        if(son[u]==-1||num[son[u]]<son[v])
            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=0;i<(int)edge[u].size();i++)
    {
        int v=edge[u][i].v;
        if(v==son[u]||v==fa[u])continue;
        getpos(v,v);
    }
}
void init()
{
    pos=0;
    memset(son,-1,sizeof(son));
    dfs(1,1,1);
    getpos(1,1);
}
void getnode(int u,int v,int d)
{
    int f1=top[u];
    int f2=top[v];
    while(f1!=f2)
    {
        if(deep[f1]<deep[f2])
        {
            swap(f1,f2);
            swap(u,v);
        }
        g1[p[f1]]+=d;
        g1[p[u]+1]-=d;
        u=fa[f1];
        f1=top[u];
    }
    if(u==v)
    {
        g1[p[u]]+=d;
        g1[p[u]+1]-=d;
        return;
    }
    if(deep[u]>deep[v])swap(u,v);
    g1[p[u]]+=d;
    g1[p[v]+1]-=d;
    return;
}
void getedge(int u,int v,int d)
{
    int f1=top[u];
    int f2=top[v];
    while(f1!=f2)
    {
        if(deep[f1]<deep[f2])
        {
            swap(f1,f2);
            swap(u,v);
        }
        g2[p[f1]]+=d;
        g2[p[u]+1]-=d;
        u=fa[f1];
        f1=top[u];
    }
    if(u==v)
        return;
    if(deep[u]>deep[v])swap(u,v);
    g2[p[son[u]]]+=d;
    g2[p[v]+1]-=d;
    return;
}
struct lede
{
    int u,v;
}e[M];
int main()
{
    int T,m,n,i,u,v,w,kk=1;
    char ch[22];
    cin>>T;
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(i=0;i<=n;i++)
            edge[i].clear();
        for(i=1;i<n;i++)
        {
            scanf("%d%d",&u,&v);
            edge[u].push_back(v);
            edge[v].push_back(u);
            e[i].u=u;
            e[i].v=v;
        }
        init();
        memset(g1,0,sizeof(g1));
        memset(g2,0,sizeof(g2));
        while(m--)
        {
            scanf("%s%d%d%d",ch,&u,&v,&w);
            if(strcmp(ch,"ADD1")==0)
                getnode(u,v,w);
            else
                getedge(u,v,w);
        }
        printf("Case #%d:\n",kk++);
        int sum=0;
        for(i=0;i<pos;i++)
        {
            sum+=g1[i];
            ans[i]=sum;
        }
        for(i=1;i<=n;i++)
        {
            if(i==1)
                printf("%d",ans[p[i]]);
            else
                printf(" %d",ans[p[i]]);
        }
        printf("\n");
        sum=0;
        for(i=1;i<pos;i++)
        {
            sum+=g2[i];
            ans[i]=sum;
        }
        for(i=1;i<n;i++)
        {
            if(deep[e[i].u]<deep[e[i].v])
                swap(e[i].u,e[i].v);
            if(i==1)
                printf("%d",ans[p[e[i].u]]);
            else
                printf(" %d",ans[p[e[i].u]]);
        }
        printf("\n");
    }
}