NOIP 模拟 $94; m 开挂$

题解 (by;zjvarphi)

最终的序列在求出来后肯定不变，就是看新产生的值是由哪个值变化而来的。

贪心策略，当两个数所生成的值交换后，差值总和不变，应当让最小的生成最大的。

直接模拟即可。

Code

#include<bits/stdc++.h>
#define ri signed
#define pd(i) ++i
#define bq(i) --i
#define func(x) std::function<x>
namespace IO{
    char buf[1<<21],*p1,*p2;
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++
    #define dg1(x) std::cerr << #x"=" << x << ' '
    #define dg2(x) std::cerr << #x"=" << x << std::endl
    #define Dg(x) assert(x)
    struct nanfeng_stream{
        template<typename T>inline nanfeng_stream &operator>>(T &x) {
            bool f=false;x=0;char ch=gc();
            while(!isdigit(ch)) f|=ch=='-',ch=gc();
            while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=gc();
            return x=f?-x:x,*this;
        }
    }cin;
}
using IO::cin;
namespace nanfeng{
    #define pb push
    #define mk std::make_pair
    #define fi first
    #define se second
    #define FI FILE *IN
    #define FO FILE *OUT
    template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
    template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
    static const int N=1e6+7;
    using ull=unsigned long long;
    int a[N],b[N],st[N],bs[N],al,cn,cnt,nm,mn,n,ct;
    // bool vis[N<<1];
    // std::vector<std::pair<int,int>> vc[N];
    std::stack<std::pair<int,int>> sta;
    std::unordered_map<int,bool> vis;
    ull ans;
    inline int main() {
        FI=freopen("openhook.in","r",stdin);
        FO=freopen("openhook.out","w",stdout);
        cin >> n;
        for (ri i(1);i<=n;pd(i)) cin >> a[i];
        for (ri i(1);i<=n;pd(i)) cin >> b[i];
        std::sort(a+1,a+n+1);
        std::sort(b+1,b+n+1);
        mn=a[1];
        for (ri i(1);i<=n;pd(i)) vis[a[i]]=true;
        a[n+1]=N+(int)2e9;
        for (ri i(1);i<=n+1;pd(i))
            if (a[i]!=a[i-1]) {
                const int tmp=a[i-1];
                nm=cmax(nm,tmp);
                if (cnt>1) {
                    for (ri j(i-cnt+1);j<i;pd(j)) {
                        while(vis.find(nm)!=vis.end()) ++nm;
                        vis[nm]=true;
                    }
                    cnt-=1;
                    int ttc=0;
                    for (ri j(tmp+1);j<a[i];pd(j)) {
                        st[++al]=j-tmp;
                        ++ttc;
                        if (ttc>=cnt) break;
                    }
                    for (ri j(tmp+ttc+1);j<cmin(a[i],nm+1);pd(j)) {
                        st[++al]=j-sta.top().fi;
                        sta.top().se-=1;
                        if (!sta.top().se) sta.pop();
                    }
                    if (cnt-ttc) sta.pb(mk(tmp,cnt-ttc));
                } else 
                    for (ri j(tmp+1);j<cmin(a[i],nm+1);pd(j)) {
                        st[++al]=j-sta.top().fi;
                        sta.top().se-=1;
                        if (!sta.top().se) sta.pop();
                    }
                if (nm<a[i]) Dg(sta.empty());
                cnt=1;
            } else ++cnt;
        std::sort(st+1,st+al+1,[](int x,int y) {return x>y;});
        for (ri i(1);i<=al;pd(i)) ans+=1ull*st[i]*b[i];
        printf("%llu
",ans);
        return 0;
    }
}
int main() {return nanfeng::main();}