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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
#530884#8551. DFS Order 5chenxinyang2006WA 32ms16224kbC++204.3kb2024-08-24 17:32:562024-08-24 17:32:57

Judging History

你现在查看的是最新测评结果

  • [2024-08-24 17:32:57]
  • 评测
  • 测评结果:WA
  • 用时:32ms
  • 内存:16224kb
  • [2024-08-24 17:32:56]
  • 提交

answer

#include <bits/stdc++.h>
#define rep(i,j,k) for(int i=(j);i<=(k);i++)
#define per(i,j,k) for(int i=(j);i>=(k);i--)
#define uint unsigned int
#define ll long long
#define ull unsigned long long
#define db double
#define ldb long double
#define pii pair<int,int>
#define pll pair<ll,ll>
#define mkp make_pair
#define eb emplace_back
#define SZ(S) (int)S.size()
//#define mod 998244353
//#define mod 1000000007
#define inf 0x3f3f3f3f
#define linf 0x3f3f3f3f3f3f3f3f
using namespace std;

template <class T>
void chkmax(T &x,T y){
	if(x < y) x = y;
}

template <class T>
void chkmin(T &x,T y){
	if(x > y) x = y;
}

inline int popcnt(int x){
	return __builtin_popcount(x);
}

inline int ctz(int x){
	return __builtin_ctz(x);
}


/*ll power(ll p,int k = mod - 2){
	ll ans = 1;
	while(k){
		if(k % 2 == 1) ans = ans * p % mod;
		p = p * p % mod;
		k /= 2;	
	}
	return ans;
}*/
int n,m;
vector <int> G[100005],son[100005];

int fa[100005],dfn[100005],siz[100005],dep[100005],ST[20][100005],Mn[20][100005],Mx[20][100005],N;
int anc(int u,int v){
	return dfn[u] <= dfn[v] && dfn[v] < dfn[u] + siz[u];
}
void dfs(int u,int f){
	fa[u] = f;
	dfn[u] = ++N;
	dep[u] = dep[f] + 1;
	siz[u] = 1;
	for(int v:G[u]){
		if(v == f) continue;
		dfs(v,u);
		son[u].eb(v);
		siz[u] += siz[v];
	}
}
int k;
int a[100005],idx[100005];
inline int cmp(int x,int y){
	if(dep[x] != dep[y]){
		if(dep[x] < dep[y]) return x;
		return y;
	}
	if(idx[x] < idx[y]) return x;
	return y;
}

inline int qry(int l,int r){
	int x = __lg(r - l + 1);
	return cmp(ST[x][l],ST[x][r - (1 << x) + 1]);
}

int chk(int u,int l,int r){
	int x = __lg(r - l + 1);
	return dfn[u] <= min(Mn[x][l],Mn[x][r - (1 << x) + 1]) && max(Mx[x][l],Mx[x][r - (1 << x) + 1]) < dfn[u] + siz[u];
}

int locate(int u,int v){
	assert(anc(u,v));
	int pos = 0;
	per(_k,16,0) if(pos + (1 << _k) < SZ(son[u]) && dfn[son[u][pos + (1 << _k)]] <= dfn[v]) pos += 1 << _k;
	return son[u][pos];
}

int concheck(int l,int r){
//	printf("concheck [%d,%d]\n",l,r);
	rep(i,l + 1,r) if(!anc(fa[a[i]],a[i - 1])) return 0;
//	printf("suc\n");
	return 1;
}

int sz;
int b[100005];
int prefcheck(int l,int r){
//	printf("prefcheck [%d,%d]\n",l,r);
	int u = a[l],p,cur = l + 1;
	sz = 0;
	while(cur <= r){
		if(sz && dep[qry(cur,r)] != dep[a[b[1]]]) break;
		p = idx[qry(cur,r)];
		b[++sz] = p;
		if(fa[a[p]] != u) return 0;
		cur = p + 1;
	}
//	rep(i,1,sz) printf("%d ",b[i]);
//	printf("\n");
	rep(i,1,sz - 1) if(!concheck(b[i],b[i + 1] - 1)) return 0;
//	printf("%d\n",b[sz]);
	if(!sz) return 1;
	return prefcheck(b[sz],r);
}

int sufcheck(int l,int r){
//	printf("sufcheck [%d,%d]\n",l,r);
	int p,cur = l;
	sz = 0;
	while(cur <= r){
		if(sz && dep[qry(cur,r)] != dep[a[b[1]]]) break;
		p = idx[qry(cur,r)];
		b[++sz] = p;
		cur = p + 1;
	}
	b[sz + 1] = r + 1;
	rep(i,1,sz) if(fa[a[b[i]]] != fa[a[b[1]]] || !concheck(b[i],b[i + 1] - 1)) return 0;
//	printf("ovo %d %d\n",b[1],l);
	if(b[1] == l) return 1;
	if(!anc(fa[a[b[1]]],a[b[1] - 1])) return 0;
	int q = locate(fa[a[b[1]]],a[b[1] - 1]);
	if(!chk(q,l,b[1] - 1)) return 0;
	return sufcheck(l,b[1] - 1);
}

int solve(){
	rep(i,1,k){
		if(idx[a[i]]) return 0;
		idx[a[i]] = i;
		ST[0][i] = a[i];
		Mn[0][i] = Mx[0][i] = dfn[a[i]];
	}
	rep(i,2,k) if(anc(a[i],a[i - 1])) return 0;
	rep(i,1,16){
		rep(j,1,k){
			if(j + (1 << i) - 1 > k) break;
			ST[i][j] = cmp(ST[i - 1][j],ST[i - 1][j + (1 << (i - 1))]);
			Mn[i][j] = min(Mn[i - 1][j],Mn[i - 1][j + (1 << (i - 1))]);
			Mx[i][j] = max(Mx[i - 1][j],Mx[i - 1][j + (1 << (i - 1))]);			
		}
	}
	sz = 0;
	int cur = 1;
	while(cur <= k){
		if(sz && dep[qry(cur,k)] != dep[a[b[1]]]) break;
		b[++sz] = idx[qry(cur,k)];
		cur = b[sz] + 1;
	}
//	rep(i,1,sz) printf("%d ",b[i]);
//	printf("\n");
	rep(i,1,sz) if(fa[a[b[i]]] != fa[a[b[1]]]) return 0;
	rep(i,1,sz - 1) if(!concheck(b[i],b[i + 1] - 1)) return 0;
	if(!prefcheck(b[sz],k)) return 0;
	if(b[1] > 1) return sufcheck(1,b[1] - 1);
	return 1;
}

int main(){	
//	freopen("test.in","r",stdin);
	scanf("%d%d",&n,&m);
	rep(i,1,n - 1){
		int u,v;
		scanf("%d%d",&u,&v);
		G[u].eb(v);G[v].eb(u);
	}
	dfs(1,0);
	rep(i,1,m){
		scanf("%d",&k);
		rep(s,1,k) scanf("%d",&a[s]);
		if(solve()) printf("Yes\n");
		else printf("No\n");
		rep(s,1,k) idx[a[s]] = 0;
	}
	return 0;
}

详细

Test #1:

score: 100
Accepted
time: 2ms
memory: 14140kb

input:

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

output:

No
No
Yes
No
No
Yes
Yes

result:

ok 7 tokens

Test #2:

score: -100
Wrong Answer
time: 32ms
memory: 16224kb

input:

10 100000
7 2
1 7
7 10
8 6
8 7
1 3
4 5
9 5
5 8
8 8 9 7 2 8 1 6 1
4 8 3 5 2
6 7 10 3 9 9 1
1 1
8 10 3 2 9 3 8 7 3
7 5 6 2 8 5 9 1
6 3 4 6 2 1 3
5 8 9 2 4 9
1 3
2 1 5
5 8 5 1 7 9
10 5 2 9 2 6 4 10 6 3 8
3 4 5 8
2 8 4
9 4 10 1 2 4 3 3 6 3
1 3
6 1 1 6 8 3 1
3 7 3 2
3 9 1 5
4 3 7 8 10
9 4 2 3 10 2 5 4 3 ...

output:

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

result:

wrong answer 20th words differ - expected: 'No', found: 'Yes'