NOIP 模拟 $31\; \rm Game$

题解

很容易求出在没有字典序最大的限制条件下的最多胜利场数。

这样就可以对于每一位放最优的解，怎么做，二分答案。

分两种情况，一种是当前一位是输的，一种是赢的，复杂度 \(\mathcal O(\rm nlog^2n)\) 卡卡常即可。

Code

#include<bits/stdc++.h>
#define ri register signed
#define p(i) ++i
namespace IO{
    char buf[1<<21],*p1=buf,*p2=buf;
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++
    struct nanfeng_stream{
        template<typename T>inline nanfeng_stream &operator>>(T &x) {
            ri f=1;x=0;register char ch=gc();
            while(!isdigit(ch)) {if (ch=='-') f=0;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 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=1e5+7;
    int an[N],bn[N],tmx[N],tx,n,mx;
    struct ZKW{
        #define ls(x) (x<<1)
        #define rs(x) (x<<1|1)
        struct segmenttree{int s,a,b;}T[N<<3];
        int bs;
        ZKW() {bs=1;}
        inline void up(int x) {
            int tmp=cmin(T[ls(x)].b,T[rs(x)].a);
            T[x].s=T[ls(x)].s+T[rs(x)].s+tmp;
            T[x].a=T[ls(x)].a+T[rs(x)].a-tmp;
            T[x].b=T[ls(x)].b+T[rs(x)].b-tmp;
        }
        inline void build() {for (;bs<=mx;bs<<=1);}
        inline void update(int p,int x,int t) {
            p+=bs;
            if (t) T[p].b+=x;
            else T[p].a+=x;
            for (p>>=1;p;p>>=1) up(p);
        }
    }T;
    inline int main() {
        //FI=freopen("nanfeng.in","r",stdin);
        //FO=freopen("nanfeng.out","w",stdout);
        cin >> n;
        for (ri i(1);i<=n;p(i)) cin >> bn[i],mx=cmax(mx,bn[i]);
        for (ri i(1);i<=n;p(i)) cin >> an[i],mx=cmax(mx,an[i]),p(tmx[an[i]]);
        tx=mx;
        T.build();
        for (ri i(1);i<=n;p(i)) T.update(bn[i],1,1),T.update(an[i],1,0);
        int ans=T.T[1].s;
        for (ri i(1);i<=n;p(i)) {
            T.update(bn[i],-1,1);
            ri l=bn[i]+1,r,res(-1);
            while(!tmx[tx]) --tx;
            r=tx;
            while(l<=r) {
                int mid(l+r>>1);
                T.update(mid,-1,0);
                if (T.T[1].s==ans-1) l=mid+1,res=mid;
                else r=mid-1;
                T.update(mid,1,0);
            }
            if (res!=-1) --ans,--tmx[res],printf("%d ",res),T.update(res,-1,0);
            else {
                l=1,r=bn[i],res;
                while(l<=r) {
                    int mid(l+r>>1);
                    T.update(mid,-1,0);
                    if (T.T[1].s==ans) l=mid+1,res=mid;
                    else r=mid-1;
                    T.update(mid,1,0);
                }
                T.update(res,-1,0);
                printf("%d ",res);
                --tmx[res];
            }
        }
        puts("");
        return 0;
    }
}
int main() {return nanfeng::main();}