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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
#310745#4996. Icy ItineraryOFforest_1273WA 4ms31328kbC++142.0kb2024-01-21 17:22:112024-01-21 17:22:12

Judging History

你现在查看的是最新测评结果

  • [2024-01-21 17:22:12]
  • 评测
  • 测评结果:WA
  • 用时:4ms
  • 内存:31328kb
  • [2024-01-21 17:22:11]
  • 提交

answer

#include<bits/stdc++.h>
#define fep(i,l,r) for(int i=l;i<=r;++i)
#define feb(i,r,l) for(int i=r;i>=l;--i)
#define For(i,u) for(int i=head[u];i;i=e[i].nxt)
#define LL long long
#define ld long double
#define pr pair<int,int>
#define mpr make_pair
using namespace std;

const int N = 1e6+5,mod = 1e9+7;
inline int read()
{
   int s=0,w=1; char ch=getchar();
   while(!(ch>='0'&&ch<='9')) {if(ch=='-') w=-1; ch=getchar();}
   while(  ch>='0'&&ch<='9')  {s=(s<<1)+(s<<3)+ch-'0'; ch=getchar();}
   return s*w;
}
inline int Mod(int x) {return x>=mod?x-mod:x;}
inline void addmod(int &x,int y) {x=Mod(x+y);}

namespace starrylasky
{
    int n,m,c0,c1,p0[N],p1[N]; vector<int > G[N];
    inline void add(int u,int v) {G[u].emplace_back(v);}
    inline void Main()
    {
        n=read(),m=read();
        fep(i,1,m)
        {
            int u=read(),v=read();
            add(u,v); add(v,u);
        }
        fep(i,1,n) sort(G[i].begin(),G[i].end());
        fep(i,1,n)
        {
            if(!c1) {p1[++c1]=i; continue;}
            if(!c0) {p0[++c0]=i; continue;}
            int u1=p1[c1],u0=p0[c0];
            int p=lower_bound(G[u1].begin(),G[u1].end(),i)-G[u1].begin();
            if(p<G[u1].size()&&G[u1][p]==i) {p1[++c1]=i; continue;}
            p=lower_bound(G[u0].begin(),G[u0].end(),i)-G[u0].begin();
            // cerr<<i<<" "<<p<<" "<<G[u0].size()<<" "<<c1<<" "<<c0<<"\n";
            if(p==G[u0].size()||(p<G[u0].size()&&G[u0][p]!=i)) {p0[++c0]=i; continue;}
            p=lower_bound(G[u1].begin(),G[u1].end(),u0)-G[u1].begin();
            if(p<G[u1].size()&&G[u1][p]==u0) --c0,p1[++c1]=u0,p1[++c1]=i;
            else --c1,p0[++c0]=u1,p0[++c0]=i;
        }
        // reverse(p1+1,p1+c1+1);
        // reverse(p0+1,p0+c0+1);
        if(p0[1]==1) swap(p0,p1),swap(c0,c1);
        fep(i,1,c1) printf("%d ",p1[i]);
        fep(i,1,c0) printf("%d ",p0[i]);
    }
}

signed main(){
    int _T=1;
    while(_T--) starrylasky::Main();
    return 0;
}/*check*/

詳細信息

Test #1:

score: 100
Accepted
time: 4ms
memory: 31328kb

input:

4 4
1 2
1 3
1 4
3 4

output:

1 3 4 2 

result:

ok qwq

Test #2:

score: 0
Accepted
time: 3ms
memory: 31004kb

input:

5 0

output:

1 2 3 4 5 

result:

ok qwq

Test #3:

score: -100
Wrong Answer
time: 3ms
memory: 30472kb

input:

10 10
7 8
7 5
5 2
6 1
10 7
4 6
5 8
3 2
10 5
1 10

output:

9 2 1 3 5 6 7 4 8 10 

result:

wrong answer The first number wasn't 1