Submission #167807

#	Time	Username	Problem	Language	Result	Execution time	Memory
167807		oolimry	Split the Attractions (IOI19_split)	C++14	100 / 100	241 ms	26360 KiB

split

#include "split.h"
#include <bits/stdc++.h>

using namespace std;

typedef vector<int> vi;
typedef pair<int,int> ii;

static vi answer;
static vi adj[100005];
static vi child[100005];
static vi tree[100005];
static int sz[100005];
static int pa[100005];
static bool vis[100005];

class UFDS {
public:
    vi p, rak, setSize;
    int numSets;

    void create(int n){
        setSize.assign(n, 1);
        numSets = n;
        rak.assign(n, 0);
        p.assign(n, 0);
        for(int i =0;i < n;i++){
            p[i] = i;
        }
    }

    int findSet(int i){
        return (p[i] == i) ? i : (p[i] = findSet(p[i]));
    }

    bool isSameSet(int i, int j){
        return findSet(i) == findSet(j);
    }

    int unionSet(int i, int j){
        if(isSameSet(i,j))
            return 0;
        numSets--;
        int x = findSet(i);
        int y = findSet(j);

        if(rak[x] > rak[y]) {
            p[y] = x; setSize[x] += setSize[y];
        }

        else {
            p[x] = y; setSize[y] += setSize[x];
            if(rak[x] == rak[y]) rak[y]++;
        }
		return max(setSize[x],setSize[y]);
    }
};

void dfs(int u){
	vis[u] = true;
	sz[u] = 1;
	for(int v : adj[u]){
		if(!vis[v]){
			child[u].push_back(v);
			tree[u].push_back(v);
			tree[v].push_back(u);
			pa[v] = u;
			dfs(v);
			sz[u] += sz[v];
		}	
	}
}

void bfs(int root, int number){
	if(number == 0) return;
	queue<int> bf;
	bf.push(root);
	while(!bf.empty()){
		int u = bf.front(); bf.pop();
		for(int v : tree[u]){
			if(answer[v] == 0){
				answer[v] = answer[u];
				number--; ///We fill only the required number of nodes
				if(number == 0) return;
				bf.push(v);
			}
		}
	}
}
vi find_split(int n, int aaa, int bbb, int ccc, vi p, vi q) {
	
	///WLOG, let A <= B <= C. A <= N/3, B <= N/2
	vector<ii> v = {ii(aaa,1),ii(bbb,2),ii(ccc,3)};
	sort(v.begin(),v.end());
	int A = v[0].first;
	int B = v[1].first;
	int AColour = v[0].second;
	int BColour = v[1].second;
	int CColour = v[2].second;
	answer.assign(n,0);
	
	for(int i = 0;i < (int) p.size();i++){
		int x = p[i], y = q[i];
		adj[y].push_back(x);
		adj[x].push_back(y);
	}
	
	///Generate an arbitrary spanning tree T
	dfs(0);
	
	///Check if there exists an edge which splits T into 2 parts which fit both A and B
	for(int i = 0;i < (int) p.size();i++){
		int x = p[i], y = q[i];
		if(sz[y] < sz[x]) swap(x,y); ///x is child of y
		
		int l = sz[x], r = n - sz[x];
		if(l > r) swap(l,r), swap(x,y);
		
		///A split is found, bfs to find nodes to put in set A and set B
		if(l >= A && r >= B){
			answer[x] = AColour;
			answer[y] = BColour;
			bfs(x,A-1); 
			bfs(y,B-1);
			for(int i = 0;i < n;i++){
				if(answer[i] == 0) answer[i] = CColour;
			}
			return answer;
		}
	}
	
	///There must exist only one vertex R such that all of it's subtrees have a size smaller than A
	int R = -1;
	for(int u = 0;u < n;u++){
		bool can = true;
		for(int v : child[u]){
			if(sz[v] >= A){
				can = false;
				break;
			}
		}
		if(n - sz[u] >= A) can = false;
		if(can){
			R = u;
			break;
		}
	}
	
	assert(R != -1);
	
	///For all edges in our spanning tree not connected to R, we merge them using UFDS
	UFDS uf;
	uf.create(n);
	for(int i = 0;i < (int) p.size();i++){
		int u = p[i]; int v = q[i];
		if(u == R || v == R) continue;
		if(pa[u] == v || pa[v] == u){
			uf.unionSet(u,v);
		}
	}

	///For all edges not in the spanning tree not connected to R, we merge them using UFDS
	///If this creates a connected component not connected to R that is at least size A, we use that for set A
	///The other set is connected via the Root R.
	for(int i = 0;i < (int) p.size();i++){
		int u = p[i]; int v = q[i];
		if(u == R || v == R) continue;
		if(pa[u] == v || pa[v] == u) continue;
			
		tree[u].push_back(v);
		tree[v].push_back(u);
		///A split is found, find the stuff to put in set A and set B using bfs
		if(uf.unionSet(u,v) >= A){ ///uf.unionSet returns the new set size of that set
			int x = u;
			int y = R;
			answer[x] = AColour;
			answer[y] = BColour;
			bfs(x,A-1);
			bfs(y,B-1);
			for(int i = 0;i < n;i++){
				if(answer[i] == 0) answer[i] = CColour;
			}
			return answer;
		}
	}
	
	///There is no answer. We have exhausted all edges and we're left with a graph with root R, and subtrees of size < A.
	///To make a set greater than A, we need to pass through R, which means we can't make sets for B.
	return answer;
}

• Subtask 1: 7 / 7
• Subtask 2: 11 / 11
• Subtask 3: 22 / 22
• Subtask 4: 24 / 24
• Subtask 5: 36 / 36