START198B——Minimize Maximum Mex

思路讲解

这个检查器还是有点难想到怎么样去写的。

auto check=[&](ll mid){
    bool oka=false,okb=false;;
    FOR(i,0,mid){
        if(cnta[i]==0){
            oka=true;
        }
        if(cntb[i]==0){
            okb=true;
        }
    }
    if(oka && okb){
        return true;
    }
    ll cnt=0;
    FOR(i,0,mid){   // 只要不是强制的，一定可以让i都在A中，或者i都在b中
        if(!forced[i]) ++cnt;
    }
    if(cnt>=2) return true;
    return false;
};

AC代码

源代码

// by Gospel_rock
#include <algorithm>
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define pb push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((long long) a.size())
#define FOR(i, a, b) for (long long i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (long long i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define lson(var) (var<<1)
#define rson(var) (var<<1|1)
#define fsp(x) fixed<<setprecision(x)

using namespace std;

using  ll = long long;using ull = unsigned long long;
using DB=double;using LD=long double;
using i128 = __int128;
using pdd = pair<DB,DB>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;using vb=vector<bool>;
using tll = tuple<ll,ll,ll>;
using vpll = vector<pll>;using vtll = vector<tll>;
using vdb = vector<DB>;using vc=vector<char>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
static constexpr ll MAXN=(ll)1e6+10,INF=(ll)1e17+9; // 1e18+9也是素数
static constexpr ll mod=(ll)1e9+7; // 998244353;
static constexpr double eps=1e-8;const double pi=acos(-1.0);
ll N,M,Q,X,K,T,lT;

inline void __();
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    
    cin>>lT;
    while(lT--)
        __();
    return 0;
}
/*

*/
inline void __(){
    cin>>N;
    vll A(N),B(N);
    set<ll> memo;
    FOR(i,0,N+5){
        memo.insert(i);
    }
    vll cnta(N+5),cntb(N+5);
    FOR(i,0,N-1){
        cin>>A[i];
        ++cnta[A[i]];
    }
    FOR(i,0,N-1){
        cin>>B[i];
        ++cntb[B[i]];
    }
    vb forced(N+5,false);
    FOR(i,0,N-1){
        if(A[i]==B[i]) forced[A[i]]=true;
    }
    auto check=[&](ll mid){
        bool oka=false,okb=false;;
        FOR(i,0,mid){
            if(cnta[i]==0){
                oka=true;
            }
            if(cntb[i]==0){
                okb=true;
            }
        }
        if(oka && okb){
            return true;
        }
        ll cnt=0;
        FOR(i,0,mid){   // 只要不是强制的，一定可以让i都在A中，或者i都在b中
            if(!forced[i]) ++cnt;
        }
        if(cnt>=2) return true;
        return false;
    };
    ll low=0,hig=N+3;
    while(low<hig){
        ll mid=low+hig>>1;
        if(check(mid)){
            hig=mid;
        }else{
            low=mid+1;
        }
    }
    cout<<low<<"\n";
}
/*
AC
https://www.codechef.com/viewsolution/1179820888
1
4
2 3 1 2
1 3 1 3
*/

心路历程（WA，TLE，MLE……）