Codeforces1158 C. Permutation recovery(拓扑排序,线段树优化建图)
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题意:
解法:
容易想到a[i]<a[nt[i]],且a[i]>a[i+1,nt[i]-1].
大的点对小的点建边,然后跑拓扑排序记录出栈时间戳dfn[],
dfn小的点赋大值即可.
1.对于a[i]<a[nt[i]],单点建边即可,
2.对于a[i]>a[i+1,nt[i]-1],需要对区间内的所有点的建边,暴力建边复杂度炸了,
用线段树优化建图即可.
总复杂度O(n*log).
code:
#include<bits/stdc++.h>
// #define SYNC_OFF
typedef std::vector<int> VE;
typedef std::pair<int,int> PI;
// #define int long long
#define ll long long
#define ull unsigned long long
//fast-coding
#define ST(x) x.begin()
#define ED(x) x.end()
#define RST(x) x.rbegin()
#define RED(x) x.end()
#define CL(x) x.clear();
#define all(a,n) a+1,a+1+n
#define ff(i,n) for(ll i=1;i<=n;i++)
#define rff(i,n) for(ll i=n;i>=1;i--)
#define fff(i,n) for(ll i=0;i<n;i++)
#define rfff(i,n) for(ll i=n-1;i>=0;i--)
#define SC(x) scanf("%s",x)
#define SL(x) strlen(x)
#define pss(a) push_back(a)
#define ps(a) push(a)
#define SZ(x) (int)x.size()
#define pee puts("");
#define eee putchar(' ');
#define re readdd()
#define pr(a) printtt(a)
int readdd(){int x=0,f=1;char c=getchar();//
while(!isdigit(c)&&c!='-')c=getchar();
if(c=='-')f=-1,c=getchar();
while(isdigit(c))x=x*10+c-'0',c=getchar();
return f*x;}
void printtt(int x){if(x<0)putchar('-'),x=-x;//
if(x>=10)printtt(x/10);putchar(x%10+'0');}
int gcd(int a,int b){return b==0?a:gcd(b,a%b);}//
int ppow(int a,int b,int mod){a%=mod;//
int ans=1%mod;while(b){if(b&1)ans=(long long)ans*a%mod;
a=(long long)a*a%mod;b>>=1;}return ans;}
bool addd(int a,int b){return a>b;}
int lowbit(int x){return x&-x;}
const int dx[4]={0,0,1,-1};
const int dy[4]={1,-1,0,0};
bool isdigit(char c){return c>='0'&&c<='9';}
bool Isprime(int x){
for(int i=2;i*i<=x;i++)if(x%i==0)return 0;
return 1;
}
void ac(int x){if(x)puts("YES");else puts("NO");}
//
using namespace std;
// const int mod=998244353;
const int mod=1e9+7;
const int maxm=2e6+5;
vector<int>g[maxm];
int pos[maxm];
int dfn[maxm];
int nt[maxm];
int d[maxm];
int a[maxm];
int n;
void add(int x,int y){
g[x].push_back(y);
d[y]++;
}
struct Tree{
int a[maxm<<2],tot;
void build(int l,int r,int node){
if(l==r){
a[node]=l;return ;
}
a[node]=++tot;
int mid=(l+r)/2;
build(l,mid,node*2);
build(mid+1,r,node*2+1);
add(a[node],a[node*2]);
add(a[node],a[node*2+1]);
}
void upd(int st,int ed,int p,int l,int r,int node){
if(st<=l&&ed>=r){
add(p,a[node]);return ;
}
int mid=(l+r)/2;
if(st<=mid)upd(st,ed,p,l,mid,node*2);
if(ed>mid)upd(st,ed,p,mid+1,r,node*2+1);
}
void init(){
ff(i,tot){
d[i]=0;
CL(g[i]);
}
tot=n;
}
}T;
bool cmp(int i,int j){
return dfn[i]>dfn[j];
}
bool topo(){
queue<int>q;
ff(i,T.tot){
if(d[i]==0){
q.push(i);
}
}
int idx=0;
while(q.size()){
int x=q.front();q.pop();
dfn[x]=++idx;
for(int &v:g[x]){
if(d[v]){
d[v]--;
if(d[v]==0){
q.push(v);
}
}
}
}
ff(i,T.tot){
if(d[i]!=0){
return 0;
}
}
ff(i,n){
pos[i]=i;
}
sort(all(pos,n),cmp);
ff(i,n){
int p=pos[i];
a[p]=i;
}
return 1;
}
void init(){
T.init();
}
void solve(){
n=re;
init();
ff(i,n)nt[i]=re;
//n+1是不存在,-1是丢失
T.build(1,n,1);
ff(i,n){
if(nt[i]==-1)continue;
int l=i+1,r=nt[i]-1;
//大的点向小的点连边
if(l<=r){
T.upd(l,r,i,1,n,1);
}
if(nt[i]!=n+1){
add(nt[i],i);
}
}
int ok=topo();
if(!ok){
pr(-1);pee;return ;
}
ff(i,n){
pr(a[i]);eee;
}
pee;
}
void Main(){
#define MULTI_CASE
#ifdef MULTI_CASE
int T;cin>>T;while(T--)
#endif
solve();
}
void Init(){
#ifdef SYNC_OFF
ios::sync_with_stdio(0);cin.tie(0);
#endif
#ifndef ONLINE_JUDGE
freopen("../in.txt","r",stdin);
freopen("../out.txt","w",stdout);
#endif
}
signed main(){
Init();
Main();
return 0;
}
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