hdu 4942

Posted 2021-02-21 microcodes

Game on S♂play

Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 942    Accepted Submission(s): 208

Problem Description

Uncle Fang is learning Splay. When he heard about the rotate operation, he would like to play a game with Captain Chen. There is a tree with n nodes initially, and the node 1 is the root. The i-th nodes gets a weight of wi, and an attribute of interesting value defined as the sum of the weight of it’s descendants. Following are two operations in this tree:

1.do a rotation on a node x. The rotation is the common rotation in splay tree, it can be right-rotation or left-rotation, as demonstrated in the following picture.If it‘s impossible to rotate, just ignore it!


2.ask Captain Chen the product of interesting values of the nodes in a subtree rooted at node x.

 

 

Input

There are several cases.The first line of the input contains a single integer T (T <= 30) which is the number of test cases.Then comes the T test cases .

For each case, the first line contains two integer numbers n and m(1<=n, m<=100000).

The following n lines describe the tree. In the i-th line, there was three number wi, xi, yi,(0<=w<=100000, 0<=x, y<=n) indicate the weight of node i, the left child of node i, and the right child of node i. If xi or yi equals to 0, it means node i has no such child.

And the following m lines describe the operations. There are two numbers p x in each line. If p is 0, it means do right-rotation on node x; If p is 1, it means do left-rotation on node x; if p is 2, it means ask the product of interesting values of the nodes in a subtree rooted at node x.

 

 

Output

For the k-th test case, first output “Case #k:” in a separate line.Then for each query,output a single num in each line after mod 1000000007.

 

 

Sample Input

1

3 4

1 2 3

1 0 0

1 0 0

2 1

0 1

2 2

2 1

 

 

Sample Output

Case #1:

3

6

2

 

 

析：给一颗n个节点的树，每个节点附带一个权值wi，而每个节点拥有兴趣值为其所有子树权值之和，要求支持三种操作：

1.对节点x右旋

2.对节点x左旋

3.查询指定节点所有子树兴趣值之积

 

说到左右旋，首先可以想到平衡树，可知平衡树左右旋仍满足中序遍历序列不变，每个节点与子树处于同一段连续序列中，故每次左右旋仅改变两个节点，其他节点不受影响，可以预处理每个节点的兴趣值以及兴趣值之积，

并依次按中序遍历顺序指定每个节点的序号，以及每个点的子树边界区间，然后用线段树维护改变的两个点上的路径即可，注意维护子树左右边界值

 

注意：模运算后，子树权值和即节点兴趣值可能为负数，故需要加上模值后再取模

 

#pragma comment (linker,"/STACK:102400000,102400000")
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#define ll long long
#define mod 1000000007
using namespace std;
const int N = 101015;
int t, n, m, ans;
int l[N], r[N], L[N], R[N], C[N], f[N];
ll w[N], s[N], sum[N], prd[N<<2];
inline void Maintain(int rt){
    L[rt] = R[rt] = C[rt];
    if(l[rt]){
        L[rt] = min(L[rt], L[l[rt]]);
        R[rt] = max(R[rt], R[l[rt]]);
    }
    if(r[rt]){
        L[rt] = min(L[rt], L[r[rt]]);
        R[rt] = max(R[rt], R[r[rt]]);
    }
}
inline ll dfs(int cur){
    s[cur] = w[cur];
    if(l[cur]){
        s[cur] += dfs(l[cur])%mod;
    }
    L[cur] = R[cur] = C[cur] = ans++;
    if(r[cur]){
        s[cur] += dfs(r[cur])%mod;
    }
    Maintain(cur);
    return sum[C[cur]] = s[cur];
}
inline void Build(int rt, int l, int r){
    if(l == r){
        prd[rt] = sum[l];
        return;
    }
    int m = (l+r)>>1;
    Build(rt<<1, l, m);
    Build(rt<<1|1, m+1, r);
    prd[rt] = prd[rt<<1]*prd[rt<<1|1]%mod;
}
inline void Modify(int rt, int l, int r, int pos, ll val){
    if(l == r){
        prd[rt] = val;
        return;
    }
    int m = (l+r)>>1;
    if(pos <= m) Modify(rt<<1, l, m, pos, val);
    else Modify(rt<<1|1, m+1, r, pos, val);
    prd[rt] = prd[rt<<1]*prd[rt<<1|1]%mod;
}
inline ll Query(int rt, int l, int r, int L, int R){
    if(L <= l && r <= R)
        return prd[rt];
    ll res = 1;
    int m = (l+r)>>1;
    if(L <= m)
        res = res*Query(rt<<1, l, m, L, R)%mod;
    if(R > m)
        res = res*Query(rt<<1|1, m+1, r, L, R)%mod;
    return res;
}

int main(){
    scanf("%d", &t);
    for (int i = 1; i <= t; i ++) {
        printf("Case #%d:
", i);
        f[0] = f[1] = 0;        //wa点，必须置零
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; i++) {
            scanf("%lld%d%d", &w[i], &l[i], &r[i]);
            f[l[i]] = i; f[r[i]] = i;
        }
        int x, y, Q;
        ans = 1, dfs(1); Build(1, 1, n);
        for(int i = 1; i <= m; i++) {
            scanf("%d%d", &Q, &x);
            if(!Q){
                y = l[x];
                if(!y) continue;
                s[x] = (s[x]+s[r[y]]-s[y]+mod)%mod; //wa点，可能出现负数
                s[y] = (s[y]+s[x]-s[r[y]]+mod)%mod;
                Modify(1, 1, n, C[x], s[x]);
                Modify(1, 1, n, C[y], s[y]);
                if(x == l[f[x]]){
                    l[f[x]] = y;
                }else{
                    r[f[x]] = y;
                }
                f[y] = f[x], l[x] = r[y], f[r[y]] = x;
                r[y] = x, f[x] = y;
                Maintain(x); Maintain(y);
            }else if(Q == 1){
                y = r[x];
                if(!y) continue;
                s[x] = (s[x]+s[l[y]]-s[y]+mod)%mod; //wa点，可能出现负数
                s[y] = (s[y]+s[x]-s[l[y]]+mod)%mod;
                Modify(1, 1, n, C[x], s[x]);
                Modify(1, 1, n, C[y], s[y]);
                if (x == l[f[x]]){
                    l[f[x]] = y;
                }else{
                    r[f[x]] = y;
                }
                f[y] = f[x], r[x] = l[y], f[l[y]] = x;
                l[y] = x; f[x] = y;
                Maintain(x); Maintain(y);
            }else {
                printf("%lld
", Query(1, 1, n, L[x], R[x]));
            }
        }
    }
}

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