Codeforces Round #517 (Div. 2) B.Curiosity Has No Limits ---- DFS

做法： DFS搜索即可，想构造写炸了……

AC代码：

#include<bits/stdc++.h>
using namespace std;

#define IO          ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define pb(x)       push_back(x)
#define sz(x)       (int)(x).size()
#define sc(x)       scanf("%d",&x)
#define abs(x)      ((x)<0 ? -(x) : x)
#define all(x)      x.begin(),x.end()
#define mk(x,y)     make_pair(x,y)
#define fin         freopen("in.txt","r",stdin)
#define fout        freopen("out.txt","w",stdout)

typedef long long ll;
typedef pair<int,int> PII;
const int mod = 1e9+7;
const double PI = 4*atan(1.0);
const int maxm = 5e6+5;
const int maxn = 1e5+5;
const int INF = 0x3f3f3f3f;
const ll LINF = 1ll<<62;

int a[maxn],b[maxn];
int ans[maxn],n;
bool flag = false;
void dfs(int x)
{
    if(x == n+1){
        flag = true;
        cout<<"YES"<<endl;
        for(int i=1;i<=n;i++) cout<<ans[i]<<" ";
        cout<<endl;
        return;
    }
    for(int i=0;i<=3;i++)
    {
        if((i|ans[x-1]) == a[x-1] && (i&ans[x-1]) == b[x-1]){
            ans[x] = i;
            dfs(x+1);
        }
    }
    return;
}
int main()
{
    // fin;
    IO;
    cin>>n;
    for(int i=1;i<n;i++) cin>>a[i];
    for(int i=1;i<n;i++) cin>>b[i];
    for(int i=0;i<=3;i++)
    {
        if(flag) break;
        ans[1] = i;
        dfs(2);
    }
    if(!flag) cout<<"NO"<<endl; 
    return 0;
}