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QOJ
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 5 | chenxinyang2006 | WA | 32ms | 16224kb | C++20 | 4.3kb | 2024-08-24 17:32:56 | 2024-08-24 17:32:57 |
Judging History
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'