CF878D D. Magic Breeding bitset

D. Magic Breeding

time limit per test

4 seconds

memory limit per test

1024 megabytes

input

standard input

output

standard output

Nikita and Sasha play a computer game where you have to breed some magical creatures. Initially, they have kcreatures numbered from 1 to k. Creatures have n different characteristics.

Sasha has a spell that allows to create a new creature from two given creatures. Each of its characteristics will be equal to the maximum of the corresponding characteristics of used creatures. Nikita has a similar spell, but in his spell, each characteristic of the new creature is equal to the minimum of the corresponding characteristics of used creatures. A new creature gets the smallest unused number.

They use their spells and are interested in some characteristics of their new creatures. Help them find out these characteristics.

Input

The first line contains integers n, k and q (1 ≤ n ≤ 105, 1 ≤ k ≤ 12, 1 ≤ q ≤ 105) — number of characteristics, creatures and queries.

Next k lines describe original creatures. The line i contains n numbers ai1, ai2, ..., ain (1 ≤ aij ≤ 109) — characteristics of the i-th creature.

Each of the next q lines contains a query. The i-th of these lines contains numbers ti, xi and yi (1 ≤ ti ≤ 3). They denote a query:

• ti = 1 means that Sasha used his spell to the creatures xi and yi.
• ti = 2 means that Nikita used his spell to the creatures xi and yi.
• ti = 3 means that they want to know the yi-th characteristic of the xi-th creature. In this case 1 ≤ yi ≤ n.

It's guaranteed that all creatures' numbers are valid, that means that they are created before any of the queries involving them.

Output

For each query with ti = 3 output the corresponding characteristic.

Examples

input

Copy

2 2 4
1 2
2 1
1 1 2
2 1 2
3 3 1
3 4 2

output

Copy

2
1

input

Copy

5 3 8
1 2 3 4 5
5 1 2 3 4
4 5 1 2 3
1 1 2
1 2 3
2 4 5
3 6 1
3 6 2
3 6 3
3 6 4
3 6 5

output

Copy

5
2
2
3
4

Note

In the first sample, Sasha makes a creature with number 3 and characteristics (2, 2). Nikita makes a creature with number 4 and characteristics (1, 1). After that they find out the first characteristic for the creature 3 and the second characteristic for the creature 4.

题目还是很好懂的吧

一开始的时候，给你k个怪兽，每个怪兽有n个属性。
然后现在你有三个操作：
第一个操作是用x和y怪物生成一个新的怪物，这个怪物的属性是x和y的最大值
第二个操作是用x和y怪物生成一个新的怪物，这个怪物的属性是x和y的最小值
第三个操作是询问第x个怪物的第y个属性是多少。

用并查集维护？不太行啊，这么多操作呢，根本无法模拟

这个bitset很有灵性

&用来取大，|用来取小

#include<bits/stdc++.h>
using namespace std;
const int N=1e6+5;
bitset<4096>S[N];
int a[12][N];
int n,k,q;
int main()
{
    scanf("%d%d%d",&n,&k,&q);
    for(int i=0; i<k; i++)
    {
        for(int j=0; j<4096; j++)
            if(j&(1<<i))S[i].set(j);
        for(int j=0; j<n; j++)scanf("%d",&a[i][j]);
    }
    int tot=k;
    while(q--)
    {
        int op,x,y;
        scanf("%d%d%d",&op,&x,&y),x--,y--;
        if(op==1)S[tot++]=S[x]&S[y];
        else if(op==2)S[tot++]=S[x]|S[y];
        else
        {
            vector<pair<int,int> >Q;
            for(int i=0; i<k; i++)Q.push_back({a[i][y],i});
            sort(Q.begin(),Q.end());
            int b = 0;
            for(int i=0; i<k; i++)
            {
                b|=1<<Q[i].second;
                if(S[x][b])
                {
                    cout<<Q[i].first<<endl;
                    break;
                }
            }
        }
    }
}