Educational Codeforces Round 136 Editorial

#	User	Rating
1	tourist	4009
2	jiangly	3823
3	Benq	3738
4	Radewoosh	3633
5	jqdai0815	3620
6	orzdevinwang	3529
7	ecnerwala	3446
8	Um_nik	3396
9	ksun48	3390
10	gamegame	3386

#	User	Contrib.
1	cry	164
1	maomao90	164
3	Um_nik	163
4	atcoder_official	160
5	-is-this-fft-	158
6	awoo	157
6	adamant	157
8	TheScrasse	154
8	nor	154
10	djm03178	153

Tutorial

Solution 1 (awoo)

for _ in range(int(input())):
	n, m = map(int, input().split())
	print(n // 2 + 1, m // 2 + 1)

Solution 2 (awoo)

#include <bits/stdc++.h>

#define forn(i, n) for (int i = 0; i < int(n); i++)

using namespace std;

int main() {
	int t;
	scanf("%d", &t);
	forn(_, t){
		int n, m;
		scanf("%d%d", &n, &m);
		int svx = 1, svy = 1;
		for (int x = 1; x <= n; ++x){
			for (int y = 1; y <= m; ++y){
				bool ok = true;
				for (int dx : {-2, -1, 1, 2}){
					for (int dy : {-2, -1, 1, 2}){
						if (abs(dx * dy) != 2) continue;
						if (1 <= x + dx && x + dx <= n && 1 <= y + dy && y + dy <= m)
							ok = false;
					}
				}
				if (ok){
					svx = x;
					svy = y;
				}
			}
		}
		printf("%d %d\n", svx, svy);
	}
}

1739B - Array Recovery

Idea: BledDest

Tutorial

Solution (Neon)

for _ in range(int(input())):
	n = int(input())
	ans = [0]
	for x in map(int, input().split()):
		if x != 0 and ans[-1] - x >= 0:
			print(-1)
			break
		else:
			ans.append(ans[-1] + x)
	else:
		print(*ans[1:])

1739C - Card Game

Idea: BledDest

Tutorial

Solution 1 (BledDest)

#include<bits/stdc++.h>

using namespace std;

long long dp[65][65][3];

void solve(int n)
{
    memset(dp, 0, sizeof dp);
    int k = n / 2;
    dp[0][0][2] = 1;
    for(int i = 0; i <= k; i++)
        for(int j = 0; j <= k; j++)
            for(int t = 0; t <= 2; t++)
            {
                int turn = (i + j) % 4;
                if (turn == 0 || turn == 3)
                {
                    if(i < k) dp[i + 1][j][t == 2 ? 0 : t] += dp[i][j][t];
                    if(j < k) dp[i][j + 1][t] += dp[i][j][t];
                }
                else
                {
                    if(i < k) dp[i + 1][j][t] += dp[i][j][t];
                    if(j < k) dp[i][j + 1][t == 2 ? 1 : t] += dp[i][j][t];
                }
            }
    for(int i = 0; i < 3; i++)
        cout << dp[k][k][i] % 998244353 << " ";
    cout << endl;
}

int main()
{
    int t;
    cin >> t;
    for(int i = 0; i < t; i++)
    {
        int n;
        cin >> n;
        solve(n);
    }
}

Solution 2 (BledDest)

def fact(n):
    return 1 if n == 0 else n * fact(n - 1)

def choose(n, k):
    return fact(n) // fact(k) // fact(n - k)

def calc(n):
    if n == 2:
        return [1, 0, 1]
    else:
        a = calc(n - 2)
        return [choose(n - 1, n // 2) + a[1], choose(n - 2, n // 2) + a[0], 1]

t = int(input())
for i in range(t):
    mod = 998244353
    n = int(input())
    a = calc(n)
    a = list(map(lambda x: x % mod, a))
    print(*a)

1739D - Reset K Edges

Idea: BledDest

Tutorial

Solution (awoo)

#include <bits/stdc++.h>

#define forn(i, n) for (int i = 0; i < int(n); i++)

using namespace std;

int n;
vector<vector<int>> g;

vector<int> st;
vector<int> pd;

void init(int v, int d){
	st.push_back(v);
	if (int(st.size()) - d >= 0)
		pd[v] = st[st.size() - d];
	for (int u : g[v])
		init(u, d);
	st.pop_back();
}

vector<char> used;

void dfs(int v){
	used[v] = true;
	for (int u : g[v]) if (!used[u])
		dfs(u);
}

int get(int d){
	pd.assign(n, -1);
	init(0, d);
	
	vector<int> ord, h(n);
	queue<int> q;
	q.push(0);
	while (!q.empty()){
		int v = q.front();
		q.pop();
		ord.push_back(v);
		for (int u : g[v]){
			q.push(u);
			h[u] = h[v] + 1;
		}
	}
	reverse(ord.begin(), ord.end());
	
	used.assign(n, 0);
	int res = 0;
	for (int v : ord) if (!used[v] && h[v] > d){
		++res;
		dfs(pd[v]);
	}
	
	return res;
}

int main() {
	int t;
	scanf("%d", &t);
	while (t--){
		int k;
		scanf("%d%d", &n, &k);
		g.assign(n, vector<int>());
		for (int i = 1; i < n; ++i){
			int p;
			scanf("%d", &p);
			--p;
			g[p].push_back(i);
		}
		int l = 1, r = n - 1;
		int ans = n;
		while (l <= r){
			int m = (l + r) / 2;
			if (get(m) <= k){
				ans = m;
				r = m - 1;
			}
			else{
				l = m + 1;
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}

1739E - Cleaning Robot

Idea: BledDest

Tutorial

Solution (awoo)

#include <bits/stdc++.h>

using namespace std;

#define forn(i, n) for(int i = 0; i < int(n); i++) 

const int INF = 1e9;

int main(){
	int n;
	cin >> n;
	vector<string> s(2);
	forn(i, 2) cin >> s[i];
	vector<array<array<int, 2>, 2>> dp(n + 1);
	forn(i, n + 1) forn(j, 2) forn(k, 2) dp[i][j][k] = -INF;
	dp[0][0][s[1][0] == '1'] = s[1][0] == '1';
	dp[0][0][0] = 0;
	forn(i, n - 1) forn(j, 2){
		int nxtj = s[j][i + 1] == '1';
		int nxtj1 = s[j ^ 1][i + 1] == '1';
		dp[i + 1][j ^ 1][0] = max(dp[i + 1][j ^ 1][0], dp[i][j][1] + nxtj1);
		dp[i + 1][j][nxtj1] = max(dp[i + 1][j][nxtj1], dp[i][j][0] + nxtj1 + nxtj);
		dp[i + 1][j][0] = max(dp[i + 1][j][0], dp[i][j][0] + nxtj);
	}
	cout << max({dp[n - 1][0][0], dp[n - 1][0][1], dp[n - 1][1][0], dp[n - 1][1][1]}) << '\n';
}

1739F - Keyboard Design

Idea: BledDest

Tutorial

Solution (BledDest)

#include<bits/stdc++.h>

using namespace std;

const int N = 10043;
const int K = 12;

int tsz = 0;
int trie[N][K];
int aut[N][K];
int lnk[N];
int p[N];
int pchar[N];
int cost[N];
int ncost[N];

int newNode()
{
    lnk[tsz] = -1;
    ncost[tsz] = -1;
    cost[tsz] = 0;
    for(int i = 0; i < K; i++)
    {
        trie[tsz][i] = aut[tsz][i] = -1;
    }
    return tsz++;
}

int nxt(int x, int y)
{                                 
    if(trie[x][y] == -1) 
    {
        trie[x][y] = newNode();
        pchar[trie[x][y]] = y;
        p[trie[x][y]] = x;
    }
    return trie[x][y];
}

int go(int x, int y);

int get_lnk(int x)
{
    if(lnk[x] != -1) return lnk[x];
    int& d = lnk[x];
    if(x == 0 || p[x] == 0) return d = 0;
    return d = go(get_lnk(p[x]), pchar[x]);    
}

int go(int x, int y)
{
    if(aut[x][y] != -1) return aut[x][y];
    int& d = aut[x][y];
    if(trie[x][y] != -1) return d = trie[x][y];
    if(x == 0) return d = 0;
    return d = go(get_lnk(x), y);
}

void add(string s, int c)
{
    int cur = 0;
    for(auto x : s) cur = nxt(cur, x - 'a');
    cost[cur] += c;     
} 

int calc(int x)
{
    if(ncost[x] != -1) return ncost[x];
    ncost[x] = cost[x];
    int y = get_lnk(x);
    if(y != x) ncost[x] += calc(y);
    return ncost[x];
}

int main()
{
    int root = newNode();
    int n;
    cin >> n;
    for(int i = 0; i < n; i++)
    {
        string s;
        int x;
        cin >> x >> s;
        map<char, set<char>> adj;
        for(int j = 0; j + 1 < s.size(); j++)
        {
            adj[s[j]].insert(s[j + 1]);
            adj[s[j + 1]].insert(s[j]);
        }
        bool bad = false;
        string res = "";
        char c;
        for(c = 'a'; c <= 'l'; c++)
        {
            if(!adj.count(c)) continue;
            if(adj[c].size() >= 3)
                bad = true;
            if(adj[c].size() == 1)
                break;
        }
        if(c == 'm' || bad) continue;
        res.push_back(c);
        while(adj[c].size() > 0)
        {
            char d = *adj[c].begin();
            adj[c].erase(d);
            adj[d].erase(c);
            c = d;
            res.push_back(c);  
        }
        bad |= adj.size() != res.size();
        map<char, int> pos;
        for(int i = 0; i < res.size(); i++)
            pos[res[i]] = i;
        for(int i = 0; i + 1 < s.size(); i++)
            bad |= abs(pos[s[i]] - pos[s[i + 1]]) > 1;
        if(bad) continue;
        add(res, x);
        reverse(res.begin(), res.end());
        add(res, x);
    }  
    int INF = 1e9;
    int K = 12;
    vector<vector<int>> dp(1 << K, vector<int>(tsz + 1, -INF));
    vector<vector<pair<int, int>>> pdp(1 << K, vector<pair<int, int>>(tsz + 1));
    dp[0][0] = 0;
    for(int i = 0; i < (1 << K); i++)
        for(int j = 0; j <= tsz; j++)
        {
            for(int z = 0; z < K; z++)
            {
                if(i & (1 << z)) continue;
                int nstate = go(j, z);
                int add = calc(nstate);
                int nmask = i | (1 << z);
                if(dp[nmask][nstate] < dp[i][j] + add)
                {
                    dp[nmask][nstate] = dp[i][j] + add;
                    pdp[nmask][nstate] = {z, j};
                }
            }
        }
    string ans = "";
    int curmask = (1 << K) - 1;
    int curstate = max_element(dp[curmask].begin(), dp[curmask].end()) - dp[curmask].begin();
    while(curmask != 0)
    {
        int cc = pdp[curmask][curstate].first;
        int ns = pdp[curmask][curstate].second;
        ans.push_back(char('a' + cc));
        curmask ^= (1 << cc);
        curstate = ns;
    }
    cout << ans << endl;
}

Comments (31)

Show archived | Write comment?

ganesh_6

2 years ago, # |

Can someone suggest problems similar to Div2 C: Card Game, which requires DP, Combinatorics and some thinking. Thanks!

→ Reply

nonrice

← Rev. 4 →

+10

For those who just want to see the recurrence used in solution for C:

$$$dp(n, p) = {n-1 \choose \frac{n}{2}}+ dp(n-2, p^{-1})$$$

Let $$$p$$$ represent some player, and $$$p^{-1}$$$ represent the opponent of that player.

adityagamer

+28

When does the robot break? Let the robot currently be in the cell (j,i) (0-indexed) and the next column with a dirty cell be nxti (possibly, nxti=i). The robot breaks only if both (1−j,nxti) and (j,nxti) are dirty.

I think there is a typo. break condition should be (1-j,nxti) and (j,nxti + 1)

leetcode07

In D, why does this greedy solution starting from the root and cutting whenever the depth exceeds the candidate does not work? TIA

Husanboy

2 years ago, # ^ |

Consider the following input

Input:

1
7 1
1 1 3 2 5 5

Output:

If you cut the tree whenever depth exceeds 2, you will require 2 operations, while the min. operations to make depth 2 is actually 1. Just make the tree and see why this happening :)

-tsukasa-

10 months ago, # ^ |

is there any way to prove that you get the smallest number of edges if you start from bottom instead of top?

xox0

Why are the constraints so low for C?
I wrote a $$$O(n)$$$ solution which should easily work for $$$\Sigma n \geq 10^6$$$

The solution

PLAn22

20 months ago, # ^ |

Can i ask a question？ why in editorial the dp[0][0][2] is 1 ？

arnavwinner

17 months ago, # ^ |

my exact same doubt.

yash_15

Someone please help with my solution to problem E:

https://mirror.codeforces.com/contest/1739/submission/174503555

idea: dp[i][j][k] ; 0<=i<=1, 0<=j<=n-1, 0<=k<=2

dp[i][j][0] = num cells to clean manually in [0-1][j..n-1] rectangle if we are currently at clean (i,j)

dp[i][j][1] = num cells to clean manually in [0-1][j..n-1] rectangle if we are currently at clean (i,j), and cell (1-i,j) is already clean

dp[i][j][2] = num cells to clean manually in [0-1][j..n-1] rectangle if we are currently at clean (i,j), and cells (1-i,j) and (1-i,j+1) are already clean

I store the index of next clean cell in row i in k00,k01,k02; and of row 1-i in k10,k11,k12

variety-jones

Take a look at Ticket 16235 from CF Stress for a counter example.

Thank you so much! Can you please clarify on the point below as well though?

Your statement looks reasonable to me. Probably a typo in the editorial.

For problem E:

" When does the robot break? Let the robot currently be in the cell $$$(j,i)$$$ (0-indexed) and the next column with a dirty cell be $$$nxt_i$$$ (possibly, $$$nxt_i=i$$$). The robot breaks only if both $$$(1−j,nxt_i)$$$ and $$$(j,nxt_i)$$$ are dirty "

Shouldn't it break when $$$(1-j,nxt_i)$$$ and $$$(j,nxt_i+1 )$$$ are both dirty, and $$$( j,nxt_i )$$$ is clean ?

Sochu

yep, that's a mistake in the editorial

PaciukZvichainyi

BledDest!
Aho-Corasick is new algo for me and and read about this in cp-algorithms. They said if you need to find all strings from a given set in text, you need to use "exit links" to make code faster:

storing the nearest leaf vertex that is reachable using suffix links (this is sometimes called the exit link).
Link

As I see, you don't use this links or ncost array does this job?

BledDest

Yeah, ncost does the trick. Usually the exit links are helpful to traverse the whole path to the root along the suffix links without actually visiting non-terminal states of the Aho-Corasick (for example, this allows to find all occurrences of all strings we have to search for in $$$O(Ans)$$$); but in this problem, we are not interested in the actual occurrences themselves, only in their total cost, so precalculating it for each state is both easier and faster.

Misuki

Can somebody explain problem E why it is not always optimal when the robot is at j-th row i-th column and $$$(1 - j, i)$$$ is dirty, $$$(j, i + 1)$$$ is clean, and we just always go to clean $$$(1 - j, i)$$$? I have been stuck by this for a long time but could not find any counter case :/

ivanzuki

+13

See this case:

8
00100110
10001101

You can see that if you clean the cell in (1, 3) (1-indexed, (row, col)) you will avoid doing some forced cleaning twice near the end.

If you are doing DP, probably you just need to add one more transition to take care of this.

Thanks you! It gets AC after I add this transition!

ezo_tihor

Can't figure out my mistake in problem D.

https://mirror.codeforces.com/contest/1739/submission/176283387

Take a look at Ticket 16293 from CF Stress for a counter example.

Thanks!

beethoven97

why my code for problem D always gives WA ? any help please .

Take a look at Ticket 16324 from CF Stress for a counter example.

Thanks a lot. it helps me to get my code accepted now. the problem is to clear visit and level arrays after each step from the binary search.

tushar19280

15 months ago, # |

can anyone help me in D ?? 219602371

15 months ago, # ^ |

Take a look at Ticket 17054 from CF Stress for a counter example.

14 months ago, # ^ |

My output is same as the expected output in the counter example. Can you help me find another counter example

It's not. Not atleast on my machine, hence the counter example. Please check for undefined behaviours in your code.

Is the submission giving 2 on your machine ?

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