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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 |
---|---|---|---|---|---|---|---|---|---|
#602862 | #8720. BFS 序 0 | liguo# | WA | 54ms | 44308kb | C++20 | 3.0kb | 2024-10-01 13:13:33 | 2024-10-01 13:13:34 |
Judging History
answer
#include<cstdio>
#include<iostream>
using namespace std;
int n,m,t;
const int maxn=2e6+10;
struct node{
int to,next;
}e[maxn];
int head[maxn],k=0;
void add(int a,int b){
e[++k].next=head[a];
e[k].to=b;
head[a]=k;
}
struct nod{
int l,r;
}bl[maxn];
int deep[maxn],h[maxn],zhan[maxn],cnt=0,fr[maxn];
void dfs(int id){
zhan[++cnt]=id;
fr[id]=cnt;
bl[cnt].l=cnt;
int tmp=cnt;
for(int i=head[id];i;i=e[i].next){
int tp=e[i].to;
deep[tp]=deep[id]+1;
dfs(tp);
}
bl[tmp].r=cnt;
}
struct ond{
int fl,mx;
}tree1[maxn*4],tree2[maxn*4];
void down(int id){
tree1[id*2].mx=tree1[id].fl;
tree1[id*2].fl=tree1[id].fl;
tree1[id*2+1].mx=tree1[id].fl;
tree1[id*2+1].fl=tree1[id].fl;
tree1[id].fl=-1;
}
void add(int id,int l,int r,int pl,int pr,int nm){
if(l==pl&&r==pr){
tree1[id].mx=nm;
tree1[id].fl=nm;
return;
}
if(tree1[id].fl>=0) down(id);
int mid=(l+r)/2;
if(pr<=mid) add(id*2,l,mid,pl,pr,nm);
else if(pl>mid) add(id*2+1,mid+1,r,pl,pr,nm);
else{
add(id*2,l,mid,pl,mid,nm);
add(id*2+1,mid+1,r,mid+1,pr,nm);
}
tree1[id].mx=max(tree1[id*2].mx,tree1[id*2+1].mx);
}
int find(int id,int l,int r,int pl,int pr){
if(l==pl&&r==pr) return tree1[id].mx;
if(tree1[id].fl>=0) down(id);
int mid=(l+r)/2;
if(pr<=mid) return find(id*2,l,mid,pl,pr);
if(pl>mid) return find(id*2+1,mid+1,r,pl,pr);
int a=find(id*2,l,mid,pl,mid);
int b=find(id*2+1,mid+1,r,mid+1,pr);
return max(a,b);
}
void downn(int id){
tree2[id*2].mx=tree2[id].fl;
tree2[id*2].fl=tree2[id].fl;
tree2[id*2+1].mx=tree2[id].fl;
tree2[id*2+1].fl=tree2[id].fl;
tree2[id].fl=-1;
}
void ad(int id,int l,int r,int pl){
tree2[id].mx++;
if(l==r) return;
if(tree2[id].fl==0) downn(id);
int mid=(l+r)/2;
if(pl<=mid) ad(id*2,l,mid,pl);
else ad(id*2+1,mid+1,r,pl);
}
int findsm(int id,int l,int r,int pl,int pr){
if(l==pl&&r==pr) return tree2[id].mx;
if(tree2[id].fl==0) downn(id);
int mid=(l+r)/2;
if(pr<=mid) return findsm(id*2,l,mid,pl,pr);
if(pl>mid) return findsm(id*2+1,mid+1,r,pl,pr);
int a=findsm(id*2,l,mid,pl,mid);
int b=findsm(id*2+1,mid+1,r,mid+1,pr);
return a+b;
}
int bk[maxn];
int main(){
scanf("%d",&n);
int a;
for(int i=2;i<=n;i++){
scanf("%d",&a);
add(a,i);
}
scanf("%d",&t);
deep[1]=1;dfs(1);
//for(int i=1;i<=n;i++) printf("%d ",deep[i]);
while(t--){
scanf("%d",&a);
for(int i=1;i<=a;i++) scanf("%d",&h[i]);
int fl=0;
for(int i=1;i<a;i++){
if(deep[h[i]]>deep[h[i+1]]){
fl=1;
break;
}
}
if(fl){
printf("No\n");
continue;
}
int kt=n;bk[n]=1;
tree1[1].mx=n;tree1[1].fl=n;
tree2[1].mx=0;tree2[1].fl=0;
for(int i=1;i<=a;i++){
int l=bl[fr[h[i]]].l,r=bl[fr[h[i]]].r;
int tmp=find(1,1,n,l,r);
int tl=bl[bk[tmp]].l,tr=bl[bk[tmp]].r;
int tsm=findsm(1,1,n,tl,tr);
//printf("%d %d %d\n",kt,tmp,tsm);
if(tmp-tsm==kt){
kt--;
bk[kt]=fr[h[i]];
ad(1,1,n,l);
add(1,1,n,l,r,kt);
}
else{
fl=1;
break;
}
}
if(fl) printf("No\n");
else printf("Yes\n");
}
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 0ms
memory: 18100kb
input:
6 1 1 3 2 4 10 4 3 6 2 5 1 4 3 2 4 5 5 2 5 4 6 3 3 1 4 2 3 5 6 3 5 4 5 2 6 1 4 4 3 2 5 4 4 6 2 3 3 3 2 6
output:
No Yes Yes No No No No No No Yes
result:
ok 10 token(s): yes count is 3, no count is 7
Test #2:
score: -100
Wrong Answer
time: 54ms
memory: 44308kb
input:
300000 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1...
output:
No No No No No No No No No No No No No No No Yes Yes No No No No No No No No No No No No No No Yes No No No Yes No No No No No No No No No Yes No No No No No No No No Yes No No Yes No No No No No No Yes No No No No No No No No Yes No No No Yes No No No No No No No Yes No No No Yes No No No No No No ...
result:
wrong answer expected YES, found NO [2nd token]