QOJ.ac

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
#236962#7513. Palindromic Beadsby_chanceWA 1ms12104kbC++142.9kb2023-11-04 12:04:482023-11-04 12:04:48

Judging History

你现在查看的是最新测评结果

  • [2024-03-27 16:34:54]
  • hack成功,自动添加数据
  • (/hack/584)
  • [2024-03-27 16:18:45]
  • hack成功,自动添加数据
  • (/hack/583)
  • [2023-11-04 12:04:48]
  • 评测
  • 测评结果:WA
  • 用时:1ms
  • 内存:12104kb
  • [2023-11-04 12:04:48]
  • 提交

answer

#include<bits/stdc++.h>
using namespace std;
const int N=2e5+5;
int n,m,pos[N][2],dp[N],ans;
struct path{int u,v,dis;}a[N];
bool operator <(const path&a,const path&b){return a.dis<b.dis;}
vector<int> G[N];
int L[N],R[N],dcnt,fa[N][20],dep[N];
void dfs(int u){
    L[u]=++dcnt;
    for(int v:G[u])if(v!=fa[u][0])
        fa[v][0]=u,dep[v]=dep[u]+1,dfs(v);
    R[u]=dcnt;
}
void pre(){
    for(int j=1;j<20;j++)
        for(int i=1;i<=n;i++)
            fa[i][j]=fa[fa[i][j-1]][j-1];
}
int LCA(int u,int v){
    if(dep[u]<dep[v])swap(u,v);int d=dep[u]-dep[v];
    for(int i=19;i>=0;i--)if((d>>i)&1)u=fa[u][i];
    if(u==v)return u;
    for(int i=19;i>=0;i--)
        if(fa[u][i]!=fa[v][i])u=fa[u][i],v=fa[v][i];
    return fa[u][0];
}
int dist(int u,int v){return dep[u]+dep[v]-2*dep[LCA(u,v)];}
int mx[N*50],ls[N*50],rs[N*50],tot;
void modify2(int p,int l,int r,int L,int R,int v){
    if(l>=L&&r<=R){mx[p]=max(mx[p],v);return;}
    int mid=(l+r)>>1;
    if(L<=mid){
        if(!ls[p])ls[p]=++tot;
        modify2(ls[p],l,mid,L,R,v);
    }
    if(R>mid){
        if(!rs[p])rs[p]=++tot;
        modify2(rs[p],mid+1,r,L,R,v);
    }
}
int query2(int p,int l,int r,int x){
    if(!p)return 0;
    if(l==r)return mx[p];
    int mid=(l+r)>>1;
    if(x<=mid)return max(mx[p],query2(ls[p],l,mid,x));
    else return max(mx[p],query2(rs[p],mid+1,r,x));
}
int root[N<<2];
void modify1(int p,int l,int r,int L,int R,int x,int y,int v){
    //if(p==1)printf("%d %d %d %d %d\n",L,R,x,y,v);
    if(l>=L&&r<=R){modify2(root[p],1,n,x,y,v);return;}
    int mid=(l+r)>>1;
    if(L<=mid)modify1(p<<1,l,mid,L,R,x,y,v);
    if(R>mid)modify1(p<<1|1,mid+1,r,L,R,x,y,v);
}
int query1(int p,int l,int r,int x,int y){
    int mx=query2(root[p],1,n,y);
    if(l==r)return mx;
    int mid=(l+r)>>1;
    if(x<=mid)return max(mx,query1(p<<1,l,mid,x,y));
    else return max(mx,query1(p<<1|1,mid+1,r,x,y));
}
int main(){
    scanf("%d",&n);
    for(int i=1,x;i<=n;i++){
        scanf("%d",&x);
        if(!pos[x][0])pos[x][0]=i;
        else pos[x][1]=i;
    }
    for(int i=1;i<=n;i++)
        if(pos[i][1])a[++m]={pos[i][0],pos[i][1],0};
    for(int i=1,u,v;i<n;i++){
        scanf("%d%d",&u,&v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    dfs(1);pre();
    for(int i=1;i<=m;i++)
        a[i].dis=dist(a[i].u,a[i].v);
    sort(a+1,a+m+1);
    for(int i=1;i<=(n<<2);i++)root[i]=++tot;
    for(int i=1;i<=m;i++){
        dp[i]=2+(a[i].dis!=1);int u=a[i].u,v=a[i].v;
        //printf(" %d %d\n",L[u],L[v]);
        dp[i]=max(dp[i],query1(1,1,n,L[u],L[v])+2);
        dp[i]=max(dp[i],query1(1,1,n,L[v],L[u])+2);
        ans=max(ans,dp[i]);
        if(L[u]>L[v])swap(u,v);
        if(L[v]>R[u])modify1(1,1,n,L[u],R[u],L[v],R[v],dp[i]);
        else{
            modify1(1,1,n,1,L[u],L[v],R[v],dp[i]);
            if(R[u]!=n)modify1(1,1,n,R[u]+1,n,L[v],R[v],dp[i]);
        }
    }
    printf("%d\n",ans);
    return 0;
}

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 12104kb

input:

4
1 1 2 2
1 2
2 3
2 4

output:

3

result:

ok single line: '3'

Test #2:

score: 0
Accepted
time: 1ms
memory: 12024kb

input:

5
1 3 2 2 1
1 2
2 3
3 4
4 5

output:

4

result:

ok single line: '4'

Test #3:

score: 0
Accepted
time: 0ms
memory: 9988kb

input:

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

output:

2

result:

ok single line: '2'

Test #4:

score: -100
Wrong Answer
time: 0ms
memory: 10068kb

input:

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

output:

0

result:

wrong answer 1st lines differ - expected: '1', found: '0'