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=1nj=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(1n,m105) .

There are  n  numbers in the next line, the  i -th means  wi(0wi109) .

Next  n1  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) (0w109)
 

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

Sample Input
   
   
   
   
6 5 1 2 3 4 5 6 1 2 1 3 2 4 2 5 4 6 2 3 5 1 5 6 2 2 3 1 1 7 2 2 4
 

Sample Output
   
   
   
   
341 348 612
 

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 long
ll 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;
}


你可能感兴趣的:(HDU,树链剖分,Naive,sometimes,LCT,5405,2015多校联合训练赛#9)