LightOJ 1348 Aladdin and the Return Journey

1348 - Aladdin and the Return Journey

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Time Limit: 2 second(s)

Memory Limit: 32 MB

Finally the Great Magical Lamp was in Aladdin's hand. Now he wanted to return home. But he didn't want to take any help from the Genie because he thought that it might be another adventure for him. All he remembered was the paths he had taken to reach there. But since he took the lamp, all the genies in the cave became angry and they were planning to attack. As Aladdin was not afraid, he wondered how many genies were there. He summoned the Genie from the lamp and asked this.

Now you are given a similar problem. For simplicity assume that, you are given a tree (a connected graph with no cycles) with n nodes, nodes represent places, edges represent roads. In each node, initially there are an arbitrary number of genies. But the numbers of genies change in time. So, you are given a tree, the number of genies in each node and several queries of two types. They are:

1)      0 i j, it means that you have to find the total number of genies in the nodes that occur in path from node i to j (0 ≤ i, j < n).

2)      1 i v, it means that number of genies in node i is changed to v (0 ≤ i < n, 0 ≤ v ≤ 1000).

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case starts with a blank line. Next line contains an integer n (2 ≤ n ≤ 30000). The next line contains n space separated integers between 0 and 1000, denoting the number of genies in the nodes respectively. Then there are n-1 lines each containing two integers: u v (0 ≤ u, v < n, u ≠ v) meaning that there is an edge from node u and v. Assume that the edges form a valid tree. Next line contains an integer q (1 ≤ q ≤ 10⁵) followed by q lines each containing a query as described above.

Output

For each case, print the case number in a single line. Then for each query 0 i j, print the total number of genies in the nodes that occur in path i to j.

Sample Input	Output for Sample Input
1 4 10 20 30 40 0 1 1 2 1 3 3 0 2 3 1 1 100 0 2 3	Case 1: 90 170

Note

Dataset is huge, use faster I/O methods.

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Problem Setter: Jane Alam Jan

 

 

这条同样是树链剖分，然后不同的是这条题是点上面有权，不是边上面有权。

然后再查询的时候注意一下dep[u]==dep[v]的时候不要直接返回，还要query一下

 

 

 

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

using namespace std;
#define root 1,n,1
#define lr rt<<1
#define rr rt<<1|1
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

typedef long long LL;
const int N= 100010;
int n,q,s;
int eh[N],et[N<<2],nxt[N<<2],w[N<<2],tot=0;
int pos,p[N],top[N],rnk[N],dep[N],fa[N],son[N],siz[N];

void init()
{
    memset(eh,-1,sizeof(eh));
    memset(son,-1,sizeof(son));
    tot=0;
    pos=1;
}

void addedge(int u,int v){
    et[tot]=v;nxt[tot]=eh[u];eh[u]=tot++;
    et[tot]=u;nxt[tot]=eh[v];eh[v]=tot++;
}

void dfs1(int u,int father,int d)
{
    dep[u]=d;
    fa[u]=father;
    siz[u]=1;
    for(int i=eh[u];~i;i=nxt[i]){
        int v=et[i];
        if(v != father){
            dfs1(v,u,d+1);
            siz[u] += siz[v];
            if( son[u]==-1 || siz[v] > siz[son[u]])
                son[u]=v;
        }
    }
}
void dfs2(int u,int tp)
{
    top[u]=tp;
    p[u]= pos++;
    rnk[ p[u] ]= u;
    if( son[u] == -1 )return ;
    dfs2(son[u], tp );
    for(int i=eh[u]; ~i ;i=nxt[i]){
        int v=et[i];
        if(v != son[u] && v!=fa[u] ){
            dfs2(v,v);
        }
    }
}

int sum[ N<<2 ];

void Up(int rt)
{
    sum[rt]=sum[lr]+sum[rr];
}

void build(int l,int r,int rt)
{
    if(l == r ){
        sum[rt]=0;
        return ;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    Up(rt);
}

void update(int l,int r,int rt,int x,int v)
{
    if( l == r ){
        sum[rt]=v;
        return ;
    }
    int m=(l+r)>>1;
    if(x<=m)
        update(lson,x,v);
    if(x>m)
        update(rson,x,v);
    Up(rt);
}
int query(int l,int r,int rt,int L,int R)
{
    int res=0;
    if( L<=l && r<=R ){
        return sum[rt];
    }
    int m=(l+r)>>1;
    if(L <= m)
        res += query(lson,L,R);
    if(R > m)
        res += query(rson,L,R);
    return res;
}

int Q(int u,int v)
{
    int res=0;
    int f1=top[u],f2=top[v];
    while(f1 != f2){
        if(dep[f1] < dep[f2]){

            swap(f1,f2);
            swap(u,v);
        }
        res += query(root,p[f1],p[u]);
        u=fa[f1];
        f1=top[u];
    }
    if(dep[u]>dep[v])swap(u,v);
    res += query(root,p[u],p[v]);
    return res;
}
int W[N];

int run()
{
    int u,v,w;
    int op;
    int _,cas=1;
    scanf("%d",&_);
    while(_--){
        printf("Case %d:
",cas++);
        scanf("%d",&n);
        init();

        for(int i=1;i<=n;++i){
            scanf("%d",&W[i]);
        }
        for(int i=0;i<n-1;++i){
            scanf("%d%d",&u,&v);
            u++;v++;
            addedge(u,v);
        }

        dfs1(1,0,0);
        dfs2( 1,1 );
        build(root);

       // for(int i=1;i<=n;++i){cout<<p[i]<<' ';}cout<<endl;

        for(int i=1;i<=n;++i)
        {
            update(root,p[i],W[i]);
        }

        scanf("%d",&q);
        while(q--)
        {
            scanf("%d",&op);
            if( !op )
            {
                scanf("%d%d",&u,&v);
                  u++;v++;
                printf("%d
",Q(u,v));
            }
            else {
                scanf("%d%d",&v,&w);
                v++;
                update(root,p[v],w);
            }
        }
    }
    return 0;
}

int main()
{
    #ifdef LOCAL
        freopen("in.txt","r",stdin);
    #endif
    ios::sync_with_stdio(0);
    return run();
}