AtCoder Beginner Contest 120 Review of past questions

Time required

Impressions

D was union-find, I tried to understand it because I tried to use it without much understanding. I want to acquire enough knowledge to be comfortable with ABC.

Problem A

Think about how many times you can listen for b yen without exceeding c.

answerA.py

a,b,c=map(int,input().split())
print(min(c,b//a))

B problem

The entire search is performed in descending order from the smallest of a and b.

answerB.py

a,b,k=map(int,input().split())
check=0
m=min(a,b)
for i in range(m,0,-1):
    if a%i==0 and b%i==0:
        check+=1
        if check==k:
            print(i)
            break

C problem

If you think that it is a problem of the pattern that considers the consistency of the number of blue and red using the stack, all of them will eventually fall, so only the cubes of the same color will remain at the end.

answerC.py

s=input()
c=0
for i in s:
    c+=(i=="0")
    c-=(i=="1")
print(len(s)-abs(c))

D problem

It was really difficult ... Or rather, the algorithm power is not enough. UnionFindTree I think it's too inexperienced to understand. For the time being, I implemented UnionFindTree by trial and error after the virtual console (I referred to the following two articles [1], [2]), [3]. First, UnionFindTree is a data structure that determines whether two elements belong to the same set and merges different sets at high speed (I will not explain how it works, so comment out in the code and See the reference article.). Also important here is that ** UnionFindTree cannot be split **. This problem requires division (because the bridge collapses) and requires some ingenuity (as mentioned in the answer, but in general ** it is difficult to delete the graph. ** ). Here, in the final state, none of the bridges are connected **, and if you perform the reverse operation, bridges will be added one by one **, so you can go back and forth when adding bridges. You can find the answer by recording the changes in the number of islands. Specifically, if you add a bridge connecting vertices a and b, the number of elements of the disjoint set that includes those vertices is $ N_a, N_b $, and $ (N_a + N_b) \ times (N_a + N_b- 1) \ div 2 -N_a \ times (N_a-1) \ div 2-N_b \ times (N_b-1) \ div 2 $ will increase the number of island pairs that can be visited. So, if you implement it with care that you want ** the number of pairs of islands that can't go back and forth ** (compared to the initial state):

answerD.cc

#include<iostream>
#include<vector>
#include<algorithm>
#include<utility>
using namespace std;
typedef long long ll;

//Reference: https://pyteyon.hatenablog.com/entry/2019/03/11/200000
//Reference: https://qiita.com/ofutonfuton/items/c17dfd33fc542c222396

class UnionFind {
public:
    vector<ll> parent; //parent[i]Is the parent of i
    vector<ll> siz; //An array representing the size of the disjoint sets(Initialize with 1)

    //Of the constructor:Behind is initializing member variables
    UnionFind(ll n):parent(n),siz(n,1){ //Initialize as everything is root at first
        for(ll i=0;i<n;i++){parent[i]=i;}
    }

    ll root(ll x){ //Recursively get the root of the tree to which the data x belongs
        if(parent[x]==x) return x;
        //The value of the assignment expression will be the value of the assigned variable!
        //Path compression(Streamline calculations by connecting elements directly to the roots)
        return parent[x]=root(parent[x]);
        //Update the parent when recursion
    }

    void unite(ll x,ll y){ //Merge x and y trees
        ll rx=root(x);//The root of x is rx
        ll ry=root(y);//y root ry
        if(rx==ry) return; //When in the same tree
        //Merge a small set into a large set(Merged from ry to rx)
        if(siz[rx]<siz[ry]) swap(rx,ry);
        siz[rx]+=siz[ry];
        parent[ry]=rx; //If x and y are not in the same tree, add y root ry to x root rx
    }

    bool same(ll x,ll y){//Returns whether the tree to which x and y belong is the same
        ll rx=root(x);
        ll ry=root(y);
        return rx==ry;
    }

    ll size(ll x){ //Disjoint size
        return siz[root(x)];
    }
};

signed main(){
    ll n,m;cin >> n >> m;
    vector< pair<ll,ll> > ab(m);
    for(ll i=m-1;i>=0;i--){cin >> ab[i].first >> ab[i].second;}
    UnionFind uf(n);
    vector<ll> ans(m);ans[0]=(n*(n-1))/2;//cout << ans[0] << endl;
    for(ll i=0;i<m-1;i++){
        //Updates are happening here at the same time
        ll a,b;a=ab[i].first-1;b=ab[i].second-1;
        if(uf.same(a,b)){
        ans[i+1]=ans[i];
        }else{
        ll sa,sb;sa=uf.size(a);sb=uf.size(b);
        ans[i+1]=ans[i]+(sa*(sa-1))/2+(sb*(sb-1))/2;
        uf.unite(a,b);sa=uf.size(a);
        ans[i+1]-=(sa*(sa-1))/2;
        }
    }
    for(ll i=m-1;i>=0;i--){
        cout << ans[i] << endl;
    }
}