QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#61820#3996. RaceltunjicWA 4ms16884kbC++141.8kb2022-11-15 03:31:272022-11-15 03:31:29

Judging History

This is the latest submission verdict.

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2022-11-15 03:31:29]
  • Judged
  • Verdict: WA
  • Time: 4ms
  • Memory: 16884kb
  • [2022-11-15 03:31:27]
  • Submitted

answer

#include <bits/stdc++.h> 

#define X first
#define Y second
#define pb push_back 

using namespace std;

typedef pair<int, int> pii;

const int LOG = 31;
const int N = 2e5 + 10;

int n, m, q, k, comp[N], base[LOG], mask[N];
int xr[N], bio[N], bioe[N], rep[N], ans[N];

vector<int> space;
vector<pii> g[N];
vector<pair<int, pii>> qs[N];

int add(int x, bool flag){
	while(x && base[__lg(x)] != -1){
		x ^= space[base[__lg(x)]];
	}
	if(!flag){
	//	printf("x = %d\n", x);
		if(x) base[__lg(x)] = space.size();
		space.pb(x);
	}
	return x;
}

void dfs(int u){
	bio[u] = true;
	for(pii e : g[u]){
		int v = e.X, c = mask[e.Y];
		if(bio[v]){
			if(!bioe[e.Y]){
				bioe[e.Y] = true;
				add(xr[u] ^ xr[v] ^ c, 0);
			}
			continue;				
		}
		bioe[e.Y] = true; 
		xr[v] = xr[u] ^ c;
		//printf("xr[%d] = %d\n", v, xr[v]);
		dfs(v);
	}
}

void rec(int u, int c){
	bio[u] = true;
	comp[u] = c;
	for(pii e : g[u]){
		if(!bio[e.X]) rec(e.X, c);
	}
}

int ne(int x){
	for(int i = 0; i < k; i++){
		x ^= 1 << i;
	}
	return x;
}

int main(){
	scanf("%d%d%d%d", &n, &m, &k, &q);
	for(int i = 0; i < m; i++){
		int u, v, c;
		scanf("%d%d%d", &u, &v, &c);
		mask[i] = 1 << (c - 1);
		g[u].pb({v, i});
		g[v].pb({u, i});
	}
	int cnt = 0;
	for(int i = 1; i <= n; i++){
		if(!bio[i]){
			rec(i, cnt);
			rep[cnt] = i;
			cnt++;
		}
	}
	memset(bio, 0, sizeof(bio));
	for(int i = 0; i < q; i++){
		int s, t;
		scanf("%d%d", &s, &t);
		if(comp[s] == comp[t]) qs[comp[s]].pb({i, {s, t}});
	}
	for(int i = 0; i < cnt; i++){
		memset(base, -1, sizeof(base));
		space.clear();
		dfs(rep[i]);
		for(pair<int, pii> p : qs[i]) ans[p.X] = min(add(xr[p.Y.X] ^ xr[p.Y.Y], 1), add(ne(xr[p.Y.X] ^ xr[p.Y.Y]), 1)) == 0;
	}
	for(int i = 0; i < q; i++){
		if(ans[i] == 0) printf("No\n");
		else printf("Yes\n");
	}
	return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 3ms
memory: 15560kb

input:

7 9 3 4
1 2 1
2 3 1
3 1 2
1 4 3
5 6 2
6 7 1
6 7 3
7 7 2
5 5 1
6 7
1 4
2 4
2 5

output:

Yes
No
Yes
No

result:

ok 4 token(s): yes count is 2, no count is 2

Test #2:

score: 0
Accepted
time: 0ms
memory: 16884kb

input:

9 10 3 5
7 2 2
4 2 1
7 1 1
2 8 1
1 7 2
2 7 3
2 8 1
7 2 1
7 5 1
8 1 2
6 4
6 7
8 4
5 1
8 6

output:

No
No
Yes
Yes
No

result:

ok 5 token(s): yes count is 2, no count is 3

Test #3:

score: -100
Wrong Answer
time: 4ms
memory: 15120kb

input:

39 30 5 999
12 22 4
11 1 1
28 13 3
35 1 4
7 17 2
20 19 4
28 7 5
15 33 5
31 38 5
13 33 4
13 35 2
16 12 5
13 33 1
21 15 3
23 32 1
19 16 3
3 22 3
11 14 2
31 26 5
32 17 5
34 17 3
31 26 2
10 37 1
1 3 1
30 12 4
1 35 3
6 1 2
25 15 2
39 23 5
10 5 3
24 34
32 27
4 13
37 28
31 36
12 11
24 6
22 32
7 17
1 15
22 ...

output:

No
No
No
No
No
Yes
No
Yes
No
No
No
No
Yes
No
No
No
Yes
No
No
No
No
No
No
No
No
No
No
Yes
No
Yes
No
Yes
Yes
Yes
No
No
No
Yes
No
No
No
No
No
Yes
No
Yes
Yes
No
No
No
No
No
Yes
No
Yes
No
No
No
No
No
No
No
No
No
Yes
No
No
No
No
Yes
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
N...

result:

wrong answer expected NO, found YES [496th token]