NOIP 模拟 $18\; \rm 老司机的狂欢$
题解 \(by\;zj\varphi\)
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 \(\rm LIS\)
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 \(\mathcal O(\rm nlognlogt+nlog^2n)\)
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\n",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\n",quea.top()),quea.pop();
}
return 0;
}
}
int main() {return nanfeng::main();}