Tree

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

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 ≤105),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, -105 ≤ k ≤ 105)

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

维护一棵树，操作是对某两点路径上的所有点的权值加上某个值，或者对某两点路径上的所有边的权值加上某个值。

分析

Kuang神出题的时候特别为这题卡了数据，LCT目测是不够过的，用树链剖分转成线性加输入挂也是勉强过，4700ms左右，那种1000~2000ms左右的算法不太明白是怎么写的。

1.维护两点路径上的点权值是树链剖分的基本操作

2.维护两点路径上的边权值需要在剖分的时候直接按照边的情况，直接用子结点的序号来代表一条边，然后做好边原本序号的记录

一般的树链剖分是直接用树状数组，但是由于本题的最终结果要求的是输出所有点和所有边的权值，因此采用树状数组每次求一次sum来得到每个点的值就有点浪费效率了。

可以参考原本树状数组的方式，直接用一个普通的数组来记录情况，在起点+value，在终点后的一个点-value，表示对整段线段完成加权值；最终求和的时候只需要从头向后扫一遍，求出每个点的sum(i)即可

/* ***********************************************
MYID    : Chen Fan
LANG    : G++
PROG    : HDU 5044
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

#define MAXN (int)1E5+10
#define MAXM (int)2E5+10

int son[MAXN],ans[MAXN];
int father[MAXN],size[MAXN],level[MAXN],data[MAXN],top[MAXN];
int start[MAXN];

typedef struct nod
{
    int to,next,no;
} node;
node edge[MAXM];
int last,n;

void addedge(int from,int to,int k)
{
    last++;
    edge[last].to=to; 
    edge[last].next=start[from];
    edge[last].no=k;
    start[from]=last;
}

int c[MAXN],pos[MAXN],fp[MAXN];
int ec[MAXN],epos[MAXN],fep[MAXN];
int tot;

int q[MAXN];
void bfs()
{
    int head=1,tail=1;
    q[1]=1;
    level[1]=0;
    father[1]=0;

    while(head<=tail)
    {
        int now=q[head];
        size[now]=1;
        for (int i=start[now];i;i=edge[i].next)
        {
            int temp=edge[i].to;
            if (temp!=father[now])
            {
                father[temp]=now;
                fep[temp]=edge[i].no;
                level[temp]=level[now]+1;
                tail++;
                q[tail]=temp;
            }
        }
        head++;
    }
    for (int i=n;i>=1;i--)
    {
        int now=q[i];
        if (father[now])
        {
            size[father[now]]+=size[now];
            if (son[father[now]]==0||size[now]>size[son[father[now]]])
            son[father[now]]=now;
        }
    }
    for (int i=1;i<=n;i++)
    {
        int now=q[i];
        if (son[father[now]]==now) top[now]=top[father[now]];
        else 
        {
            top[now]=now;
            while(now)
            {
                tot++;
                pos[now]=tot;
                fp[tot]=now;
                now=son[now];
            }
        }
    }
}

void change(int x,int y,int value)
{
    while(top[x]!=top[y])
    {
        if (level[top[x]]<level[top[y]]) swap(x,y);
    
        c[pos[top[x]]]+=value;
        c[pos[x]+1]-=value;

        x=father[top[x]];
    }
    if (level[x]>level[y]) swap(x,y);
    
    c[pos[x]]+=value;
    c[pos[y]+1]-=value;
}

void change_edge(int x,int y,int value)
{
    while(top[x]!=top[y])
    {
        if (level[top[x]]<level[top[y]]) swap(x,y);
    
        ec[pos[top[x]]]+=value;
        ec[pos[x]+1]-=value;

        x=father[top[x]];
    }
    if (level[x]>level[y]) swap(x,y);
    
    ec[pos[son[x]]]+=value;
    ec[pos[y]+1]-=value;
}

int INT() {
    char ch;
    int res;
    bool neg;
    while (ch = getchar(), !isdigit(ch) && ch != '-')
        ;
    if (ch == '-') {
        neg = true;
        res = 0;
    } else {
        neg = false;
        res = ch - '0';
    }
    while (ch = getchar(), isdigit(ch))
        res = res * 10 + ch - '0';
    return neg ? -res : res;
}

int main()
{
    freopen("5044.txt","r",stdin);

    int t;
    t=INT();

    for (int tt=1;tt<=t;tt++)
    {
        printf("Case #%d:\n",tt);

        int m;
        n=INT();m=INT();
    
        memset(start,0,sizeof(start));
        last=0;
        for (int i=1;i<n;i++)
        {
            int u,v;
            u=INT();
            v=INT();
            addedge(u,v,i);
            addedge(v,u,i);
        }

        memset(son,0,sizeof(son));
        tot=0;
        bfs();

        memset(c,0,sizeof(c));
        memset(ec,0,sizeof(ec));

        for (int i=1;i<=m;i++)
        {
            char s;
            int c1,c2,k;
            s=getchar();
            while (!isdigit(s)) s=getchar();
            c1=INT();
            c2=INT();
            k=INT();
            if (s=='1') change(c1,c2,k);
            else change_edge(c1,c2,k);
        }

        int ss=0;
        for (int i=1;i<=n;i++)
        {
            ss+=c[i];
            ans[fp[i]]=ss;
        }
        printf("%d",ans[1]);
        for (int i=2;i<=n;i++) 
        {
            putchar(' ');
            printf("%d",ans[i]);
        }
        putchar('\n');

        ss=0;
        for (int i=1;i<=n;i++)
        {
            ss+=ec[i];
            ans[fep[fp[i]]]=ss;
        }
        if (n>1) printf("%d",ans[1]);
        for (int i=2;i<n;i++) 
        {
            putchar(' ');
            printf("%d",ans[i]);
        }
        putchar('\n');
        
    }

    return 0;
}

Chenfan Blog

HDU 5044 Tree 树链剖分