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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#64072#3996. RaceltunjicWA 2ms13004kbC++141.9kb2022-11-24 05:11:462022-11-24 05:11:47

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.
  • [2022-11-24 05:11:47]
  • Judged
  • Verdict: WA
  • Time: 2ms
  • Memory: 13004kb
  • [2022-11-24 05:11:46]
  • 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 N = 2e5 + 10;
const int LOG = 30;

int n, m, q, k, dep[N], xr[N], base[LOG], comp[N];
int ans[N];
bool vis[N];
pii task[N];

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

bool add(int x, bool flag){
	while(x && base[__lg(x)] != -1){
		x ^= space[base[__lg(x)]];
	}  
	if(flag && x){
		base[__lg(x)] = space.size();
		space.pb(x);
	}
	return x == 0;
}

void dfs(int u, int c){
	comp[u] = c;
	for(pii e : g[u]){
		int v = e.X, val = e.Y;
		if(comp[v]){
			continue;
		}
		dep[v] = dep[u] + 1;
		xr[v] = xr[u] ^ (1 << val);
		dfs(v, c);
	}
}

void solve(int u){
	for(pii e : g[u]){
		int v = e.X, val = e.Y;
		vis[val] = true;
		if(dep[v] <= dep[u]){
			bool res = add(xr[u] ^ xr[v] ^ (1 << val), true);
			continue;
		} else if(dep[v] == dep[u] + 1)
			solve(v);
	}
}

int main(){
	memset(base, -1, sizeof(base));
	scanf("%d%d%d%d", &n, &m, &k, &q);
	for(int i = 0; i < m; i++){
		int u, v, val;
		scanf("%d%d%d", &u, &v, &val);
		g[u].pb({v, val - 1});
		g[v].pb({u, val - 1});
	}
	for(int i = 1; i <= n; i++){
		if(comp[i]) continue;
		dfs(i, i);
	}
	for(int i = 0; i < q; i++){
		int u, v;
		scanf("%d%d", &u, &v);
		task[i] = {u, v};
		if(comp[u] == comp[v]){
			t[comp[u]].pb(i);
		}	
	}
	for(int i = 1; i <= n; i++){
		if(t[i].empty()) continue; 
		memset(base, -1, sizeof(base));
		memset(vis, 0, sizeof(vis));
		space.clear();
		solve(i);
		bool all = true;
		for(int j = 0; j < k; j++) all &= vis[j]; 
		for(int ind : t[i]){
			int u = task[ind].X;
			int v = task[ind].Y;
			int val = xr[u] ^ xr[v];
			if(u == v){
				ans[ind] = true;
				continue;
			}
			if(add(val, false) || add(((1 << k) - 1) ^ val, false)) ans[ind] = all & true;
		} 
		break;
	}
	for(int i = 0; i < q; i++){
		printf(ans[i] ? "Yes\n" : "No\n");
	}
	return 0;
}

详细

Test #1:

score: 0
Wrong Answer
time: 2ms
memory: 13004kb

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:

No
No
Yes
No

result:

wrong answer expected YES, found NO [1st token]