X day4

题目

官方题解

T1:

单调栈，单调队列因为认为考场上会写崩所以写了一个十分暴力的方法(线段树)

然后做一做区间覆盖即可

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
#define int long long
using namespace std;
inline int read(){
    int f=,ans=;char c;
    while(c<''||c>''){if(c=='-')f=-;c=getchar();}
    while(c>=''&&c<=''){ans=ans*+c-'';c=getchar();}
    return f*ans;
}
int k[],n,a[],tmp[],res[],ans[];
void pushdown(int k,int l,int r){
    if(res[k]==-) return;
    ans[k<<]=res[k];
    ans[k<<|]=res[k];
    res[k<<]=res[k<<|]=res[k];
    res[k]=-;
    return;
}
int query1(int k,int l,int r,int x,int y){
    if(x>y) return ;
    if(x<=l&&r<=y) return ans[k];
    int mid=l+r>>,maxn=;
    pushdown(k,l,r);
    if(x<=mid) maxn=max(maxn,query1(k<<,l,mid,x,y));
    if(mid<y) maxn=max(maxn,query1(k<<|,mid+,r,x,y));
    ans[k]=max(ans[k<<],ans[k<<|]);
    return maxn;
}
int query(int k,int l,int r,int x,int y){

    if(x>y) return <<-;
    if(x<=l&&r<=y) {
        return ans[k];
    }
    pushdown(k,l,r);
    int mid=l+r>>,minn=<<-;
    if(x<=mid) minn=min(minn,query(k<<,l,mid,x,y));
    if(mid<y) minn=min(minn,query(k<<|,mid+,r,x,y));
    ans[k]=min(ans[k<<],ans[k<<|]);
    return minn;
}
void update1(int k,int l,int r,int x,int y,int pos){
    if(x>y) return;
    if(x<=l&&r<=y){ans[k]=res[k]=pos;return;}
    pushdown(k,l,r);
    int mid=l+r>>;
    if(x<=mid) update1(k<<,l,mid,x,y,pos);
    if(mid<y) update1(k<<|,mid+,r,x,y,pos);
    ans[k]=max(ans[k<<],ans[k<<|]);
    return;
}
void update(int k,int l,int r,int x,int y,int pos){
    if(x>y) return;
    if(x<=l&&r<=y){ans[k]=res[k]=pos;return;}
    pushdown(k,l,r);
    int mid=l+r>>;
    if(x<=mid) update(k<<,l,mid,x,y,pos);
    if(mid<y) update(k<<|,mid+,r,x,y,pos);
    ans[k]=min(ans[k<<],ans[k<<|]);
    return;
}
int h[],anss[],maxv;
signed main(){
    memset(res,-,sizeof(res));
    n=read();
    for(int i=;i<=n;i++) k[i]=tmp[i]=a[i]=read(),h[i]=read();
    sort(tmp+,tmp+n+);
    for(int i=;i<=n;i++) a[i]=lower_bound(tmp+,tmp+n+,a[i])-tmp;
    for(int i=;i<=n;i++){
        int t1=query1(,,n,a[i]+,n);
        anss[t1]+=h[i];
        update1(,,n,,a[i],i);
    }
    memset(res,-,sizeof(res));
    memset(ans,/,sizeof(ans));
    for(int i=n;i>=;i--){
        int t1=query(,,n,a[i]+,n);
        if(t1>=&&t1<=n) anss[t1]+=h[i];
        update(,,n,,a[i],i);
    }
    for(int i=;i<=n;i++)maxv=max(maxv,anss[i]);
    cout<<maxv;
}

T2:

易发现答案是具有单调性的，所以二分判一判即可

时间复杂度:$O(n^2 \log n)$

// luogu-judger-enable-o2
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
inline int read(){
    int f=,ans=;char c;
    while(c<''||c>''){if(c=='-')f=-;c=getchar();}
    while(c>=''&&c<=''){ans=ans*+c-'';c=getchar();}
    return f*ans;
}
int c,n;
int x[],y[],minn=<<-,xp[];
struct node{
    int x,y;
}st[];
bool CHECK(int l,int r,int w){
    if(r-l+<c) return ;
    int cnt=;
    int yp[];
    memset(yp,,sizeof(yp));
    for(int i=l;i<=r;i++) yp[++yp[]]=st[i].y;
    sort(yp+,yp+yp[]+);
    for(int i=c;i<=yp[];i++){
        if(yp[i]-yp[i-c+]<=w) {
            return ;
        }
    }
    return ;
}
bool cmp(node x1,node x2){return x1.x<x2.x;}
bool check(int xx){
    int l=;
    for(int i=;i<=n;i++) st[i].x=x[i],st[i].y=y[i];
    sort(st+,st+n+,cmp);
    for(int r=;r<=n;r++){
        if(st[r].x-st[l].x>xx){
            if(CHECK(l,r-,xx)) {
                return ;
            }
            while(st[r].x-st[l].x>xx) l++;
        }
    }
    if(CHECK(l,n,xx)) return ;
    return ;
}
int main(){
    c=read(),n=read();
    for(int i=;i<=n;i++) x[i]=read(),y[i]=read();
    int l=,r=;
    while(l<=r){
        int mid=l+r>>;
        if(check(mid)) minn=min(minn,mid),r=mid-;
        else l=mid+;
    }
    cout<<minn+;
}