NOIP 模拟 $18; m 老司机的狂欢$

题解 (by;zjvarphi)

一道很有趣的题，我用的动态开点线段树和倍增

首先对于第一问，不难想到要二分，二分时间，因为时间长一定不会比时间短能跑的人多

那么如何 check，先将所有老司机按初始坐标排个序，之后将每个老司机跑的距离加上，那么就是对之后的序列求一个 ( m LIS)

求这个长度可以用树装数组维护值域，而求这个字典序最小的序列就需要用到倍增；类似于树上倍增，对于 (i,j) 的序列，他们在相同的一位之前一定是一样的，而之后哪个的最小值小，那个就更优

我们对于序列长度开一棵动态开点线段树，每个节点存的是这一段位置的最优选择，每次回溯时维护即可

复杂度就是 (mathcal O( m nlognlogt+nlog^2n))

Code

#include<bits/stdc++.h>
#define ri register signed
#define p(i) ++i
using namespace std;
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)?EOF:*p1++
    template<typename T>inline void read(T &x) {
        ri f=1;x=0;register char ch=gc();
        while(ch<'0'||ch>'9') {if (ch=='-') f=0;ch=gc();}
        while(ch>='0'&&ch<='9') {x=(x<<1)+(x<<3)+(ch^48);ch=gc();}
        x=f?x:-x;
    }
}
using IO::read;
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;}
    typedef long long ll;
    typedef double db;
    static const int N=1e5+7;
    struct node{ll pos;int a,id;}dr[N];
    int p[N],pi[N],pre[N][21],mn[N][21],tpr[N],dp[N],num,n,k,ed;
    db pos[N],tmp[N];
    priority_queue<int,vector<int>,greater<int> > quea;
    inline int operator<(const node &n1,const node &n2) {return n1.pos<n2.pos;}
    inline int cmp(int x,int y) {return pos[x]<pos[y];}
    struct BIT{
        #define lowbit(x) ((x)&-(x))
        int c[N];
        inline void update(int x,int len) {
            for (ri i(x);i<=n;i+=lowbit(i)) c[i]=cmax(c[i],len);
        }
        inline int query(int x) {
            int res=0;
            if (!x) return res;
            for (ri i(x);i;i-=lowbit(i)) res=cmax(res,c[i]);
            return res;
        }
    }B;
    struct Seg{
        #define ls(x) T[x].l
        #define rs(x) T[x].r
        struct segmenttree{int l,r,cur;}T[N*21];
        int rt[N],tot;
        inline int calc(int x1,int x2) {
            ri mn1=INT_MAX,mn2=INT_MAX;
            ri tx1=x1,tx2=x2;
            for (ri i(17);~i;--i) 
                if (pre[x1][i]!=pre[x2][i]) {
                    mn1=cmin(mn1,mn[x1][i]),mn2=cmin(mn2,mn[x2][i]);
                    x1=pre[x1][i],x2=pre[x2][i];
                }
            mn1=cmin(mn1,mn[x1][0]),mn2=cmin(mn2,mn[x2][0]);
            if (mn1<mn2) return tx1;
            else return tx2;
        } 
        inline void up(int x) {
            if (!ls(x)) T[x].cur=T[rs(x)].cur;
            else if (!rs(x)) T[x].cur=T[ls(x)].cur;
            else {
                ri x1=T[ls(x)].cur,x2=T[rs(x)].cur;
                T[x].cur=calc(x1,x2);
            }
        }
        void update(int &x,int k,int p,int l,int r) {
            if (!x) x=p(tot);
            if (l==r) return (void)(T[x].cur=k);
            int mid(l+r>>1);
            if (p<=mid) update(ls(x),k,p,l,mid);
            else update(rs(x),k,p,mid+1,r);
            up(x);
        }
        int query(int x,int l,int r,int lt,int rt) {
            if (!x) return 0;
            if (l<=lt&&rt<=r) return T[x].cur;
            int mid(lt+rt>>1),tmp1(0),tmp2(0);
            if (l<=mid) tmp1=query(ls(x),l,r,lt,mid);
            if (r>mid) tmp2=query(rs(x),l,r,mid+1,rt);
            if (!tmp1||!tmp2) return (tmp1|tmp2);
            return calc(tmp1,tmp2);
        }
        int tst(int x,int p,int l,int r) {
            if (!x) return 0;
            if (l==r) return T[x].cur;
            int mid(l+r>>1);
            if (p<=mid) return tst(ls(x),p,l,mid); 
            else return tst(rs(x),p,mid+1,r);
        }
    }T;
    inline int check(int tim) {
        num=0;
        memset(B.c,0,sizeof(B.c));
        memset(dp,0,sizeof(dp));
        for (ri i(1);i<=n;p(i)) pos[i]=0.5*(db)dr[i].a*tim*(db)tim+(db)dr[i].pos,tmp[i]=pos[i];
        sort(tmp+1,tmp+n+1);
        ri kk=unique(tmp+1,tmp+n+1)-tmp;
        for (ri i(1);i<=n;p(i)) pi[i]=lower_bound(tmp+1,tmp+kk,pos[i])-tmp;
        for (ri i(1);i<=n;p(i)) {
            ri cur=p[i],curp=pi[i];
            int res=B.query(curp-1);
            if (res+1>num) num=res+1;
            dp[i]=res+1;
            B.update(curp,res+1);
        }
        if (num>=k) return 1;
        return 0;
    }
    inline int main() {
        // FI=freopen("nanfeng.in","r",stdin);
        // FO=freopen("nanfeng.out","w",stdout);
        read(n),read(k);
        for (ri i(1);i<=n;p(i)) read(dr[i].pos),read(dr[i].a),dr[i].id=i;
        sort(dr+1,dr+n+1);
        for (ri i(1);i<=n;p(i)) p[i]=dr[i].id;
        ri l=0,r=86400,ans;
        while(l<=r) {
            int mid(l+r>>1);
            if (check(mid)) l=mid+1,ans=mid;
            else r=mid-1;
        }
        check(ans);
        printf("%d
",ans);
        if (num>k) puts("-1");
        else {
            memset(mn,0x3f,sizeof(mn));
            for (ri i(1);i<=n;p(i)) {
                ri cur=p[i];
                mn[cur][0]=cur;
                if (dp[i]-1>0) {
                    pre[cur][0]=T.query(T.rt[dp[i]-1],1,pi[i]-1,1,n);
                    for (ri j(1);j<=17;p(j)) {
                        mn[cur][j]=cmin(mn[cur][j-1],mn[pre[cur][j-1]][j-1]);
                        pre[cur][j]=pre[pre[cur][j-1]][j-1];
                    }
                }
                T.update(T.rt[dp[i]],cur,pi[i],1,n);
            }
            ri cur=T.query(T.rt[num],1,n,1,n); 
            while(pre[cur][0]) quea.push(cur),cur=pre[cur][0];                
            quea.push(cur);
            while(!quea.empty()) printf("%d
",quea.top()),quea.pop();    
        }
        return 0;
    }  
}
int main() {return nanfeng::main();}