HDU 5405 Sometimes Naive 2015多校联合训练赛#9 LCT 树链剖分

Sometimes Naive

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 11    Accepted Submission(s): 6

Problem Description

Rhason Cheung had a naive problem, and asked Teacher Mai for help. But Teacher Mai thought this problem was too simple, sometimes naive. So she ask you for help.

She has a tree with n vertices, numbered from 1 to n. The weight of i-th node is wi.

You need to support two kinds of operations: modification and query.

For a modification operation u,w, you need to change the weight of u-th node into w.

For a query operation u,v, you should output ∑ni=1∑nj=1f(i,j). If there is a vertex on the path from u to v and the path from i to j in the tree, f(i,j)=wiwj, otherwise f(i,j)=0. The number can be large, so print the number modulo 109+7

 

Input

There are multiple test cases.

For each test case, the first line contains two numbers n,m(1≤n,m≤105).

There are n numbers in the next line, the i-th means wi(0≤wi≤109).

Next n−1 lines contain two numbers each, ui and vi, that means that there is an edge between ui and vi.

The following are m lines. Each line indicates an operation, and the format is "1 u w"(modification) or "2 u v"(query)(0≤w≤109)

 

Output

For each test case, print the answer for each query operation.

 

Sample Input

6 51 2 3 4 5 61 21 32 42 54 62 3 51 5 62 2 31 1 72 2 4

 

Sample Output

341348612

 

Author

xudyh

 

Source

2015 Multi-University Training Contest 9

根据题意求所有i,j路径经过路径u,v上的点，那么对wi*wj累加

反面考虑：所有路径累加和 - 不经过路径的累加和

那么只需快速得到不在路径上子树的规模，他们的平方的累加和就是结果。

解题报告用的是树链剖分。我还没想好怎么实现。

比赛的时候想到LCT了，但是之前没写过用子树标记父亲的标记的代码，所以最后也没打出来。

但是之前看过别人有实现过，晚上想一想也想到了。

如何实现LCT的结点记录自己子树的信息：

        用一个标记如S1，记录，这个标记只有在断开连接或者建立连接的时候才会更新。然后用一个S2，这个标记是用于splay上做更新的

于是s1的更改只需在access操作更改，而s2只需update时更新即可。

如果要用我的代码当模板注意我的注释哦。

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<vector>using namespace std;#define maxn 200007#define inf  1000000000#define ll long longll mod = 1000000007;struct Node{    Node *fa,*ch[2];    bool rev,root;//基本结构不需要修改    ll val,s1,s2,s3,s4;//这里的标记可以更改};Node pool[maxn];Node *nil,*tree[maxn];int cnt = 0;void init(){    cnt = 1;    nil = tree[0] = pool;    nil->ch[0] = nil->ch[1] = nil;    nil->s1 = nil->s2 = nil->val = nil->s3 = nil->s4 = 0;}Node *newnode(ll val,Node *f){     pool[cnt].fa = f;    pool[cnt].ch[0]=pool[cnt].ch[1]=nil;    pool[cnt].rev = false;    pool[cnt].root = true;//以下可修改    pool[cnt].val = val;    pool[cnt].s1 = 0;    pool[cnt].s2 = 0;    pool[cnt].s3 = 0;    pool[cnt].s4 = 0;    return &pool[cnt++];}//左右子树反转******真正把结点变为根void update_rev(Node *x){    if(x == nil) return ;    x->rev = !x->rev;    swap(x->ch[0],x->ch[1]);}//splay向上更新信息******void update(Node *x){//splay需要自己写更新代码    if(x == nil) return ;    x->s2 = x->s1 + x->val;    x->s4 = x->s3;    if(x->ch[0] != nil){        x->s2 += x->ch[0]->s2;        if(x->s2 >= mod)            x->s2 -= mod;        x->s4 += x->ch[0]->s4;        x->s4 %= mod;    }    if(x->ch[1] != nil){        x->s2 += x->ch[1]->s2;        if(x->s2 >= mod)            x->s2 -= mod;        x->s4 += x->ch[1]->s4;        x->s4 %= mod;    }}//splay下推信息******void pushdown(Node *x){//树的反转，才能保证把一个结点提为根。如果有其他下推操作，自己加入    if(x->rev != false){        update_rev(x->ch[0]);        update_rev(x->ch[1]);        x->rev = false;    }}//splay在root-->x的路径下推信息******void push(Node *x){//不需要改    if(!x->root) push(x->fa);    pushdown(x);}//将结点x旋转至splay中父亲的位置******void rotate(Node *x){//旋转操作，不太需要改，除非有懒标记，需要加上pushdown，update    Node *f = x->fa, *ff = f->fa;    int t = (f->ch[1] == x);    if(f->root)        x->root = true, f->root = false;    else ff->ch[ff->ch[1] == f] = x;    x->fa = ff;    f->ch[t] = x->ch[t^1];    x->ch[t^1]->fa = f;    x->ch[t^1] = f;    f->fa = x;    update(f);}//将结点x旋转至x所在splay的根位置******void splay(Node *x){//不需要改    push(x);    Node *f, *ff;    while(!x->root){        f = x->fa,ff = f->fa;        if(!f->root)            if((ff->ch[1]==f)&&(f->ch[1] == x)) rotate(f);            else rotate(x);        rotate(x);    }    update(x);}//将x到树根的路径并成一条path******Node *access(Node *x){    Node *y = nil,*z;    while(x != nil){        splay(x);        z = x->ch[1];//记录断开的结点        x->ch[1]->root = true;        (x->ch[1] = y)->root = false;        update(z);        x->s1 += z->s2;//更新父亲信息        x->s3 += (z->s2*z->s2);        x->s1 %= mod;        x->s3 %= mod;        x->s1 -= y->s2;        x->s3 -= y->s2*y->s2;        x->s1 %= mod;        x->s3 %= mod;        update(x);        y = x;        x = x->fa;    }    return y;}//将结点x变成树根******void be_root(Node *x){//基本操作    access(x);    splay(x);    update_rev(x);}int value[maxn];vector<int> head[maxn];void dfs(int u,int f){//dfs建立没有链的森林    tree[u]->fa = tree[f];    for(int i = 0;i < head[u].size() ;i++){        int v = head[u][i];        if(v == f) continue;        dfs(v,u);        tree[u]->s1 += tree[v]->s2;        tree[u]->s1 %= mod;        tree[u]->s3 += tree[v]->s2*tree[v]->s2;        tree[u]->s3 %= mod;    }    update(tree[u]);}int main(){    int n,m;    while(scanf("%d%d",&n,&m)!=EOF){        for(int i = 1;i <= n; i++)            scanf("%d",&value[i]);        for(int i = 0;i <= n; i++)            head[i].clear();        int u,v;        for(int i = 1;i < n; i++){            scanf("%d%d",&v,&u);            head[v].push_back(u);            head[u].push_back(v);        }        init();        for(int i=1;i <= n;i++)            tree[i] = newnode(value[i],nil);        dfs(1,0);        ll total = 0;        for(int i = 1;i <= n; i++)            total += value[i];        total %= mod;        int t;        for(int i = 0;i < m; i++){            scanf("%d%d%d",&t,&u,&v);            if(t == 1){                be_root(tree[u]);                access(tree[u]);                total = total + v - tree[u]->val;                total %= mod;                tree[u]->val = v;                update(tree[u]);            }            else {                be_root(tree[u]);                access(tree[v]);                splay(tree[v]);                ll ans = total*total%mod-tree[v]->s4;                ans = (ans%mod+mod)%mod;                printf("%I64d\n",ans);            }        }    }    return 0;}