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=1,ans=0;char c;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();}
    return f*ans;
}
int k[800001],n,a[800001],tmp[800001],res[800001],ans[800001];
void pushdown(int k,int l,int r){
    if(res[k]==-1) return;
    ans[k<<1]=res[k];
    ans[k<<1|1]=res[k];
    res[k<<1]=res[k<<1|1]=res[k];
    res[k]=-1;
    return;
}
int query1(int k,int l,int r,int x,int y){
    if(x>y) return 0;
    if(x<=l&&r<=y) return ans[k];
    int mid=l+r>>1,maxn=0;
    pushdown(k,l,r);
    if(x<=mid) maxn=max(maxn,query1(k<<1,l,mid,x,y));
    if(mid<y) maxn=max(maxn,query1(k<<1|1,mid+1,r,x,y));
    ans[k]=max(ans[k<<1],ans[k<<1|1]);
    return maxn;
}
int query(int k,int l,int r,int x,int y){
    
    if(x>y) return 2<<30-1;
    if(x<=l&&r<=y) {
        return ans[k];
    }
    pushdown(k,l,r);
    int mid=l+r>>1,minn=2<<30-1;
    if(x<=mid) minn=min(minn,query(k<<1,l,mid,x,y));
    if(mid<y) minn=min(minn,query(k<<1|1,mid+1,r,x,y));
    ans[k]=min(ans[k<<1],ans[k<<1|1]);
    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>>1;
    if(x<=mid) update1(k<<1,l,mid,x,y,pos);
    if(mid<y) update1(k<<1|1,mid+1,r,x,y,pos);
    ans[k]=max(ans[k<<1],ans[k<<1|1]);
    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>>1;
    if(x<=mid) update(k<<1,l,mid,x,y,pos);
    if(mid<y) update(k<<1|1,mid+1,r,x,y,pos);
    ans[k]=min(ans[k<<1],ans[k<<1|1]);
    return;
}
int h[50001],anss[50001],maxv;
signed main(){
    memset(res,-1,sizeof(res));
    n=read();
    for(int i=1;i<=n;i++) k[i]=tmp[i]=a[i]=read(),h[i]=read();
    sort(tmp+1,tmp+n+1);
    for(int i=1;i<=n;i++) a[i]=lower_bound(tmp+1,tmp+n+1,a[i])-tmp;
    for(int i=1;i<=n;i++){
        int t1=query1(1,1,n,a[i]+1,n);
        anss[t1]+=h[i];
        update1(1,1,n,1,a[i],i);
    }
    memset(res,-1,sizeof(res));
    memset(ans,127/3,sizeof(ans));
    for(int i=n;i>=1;i--){
        int t1=query(1,1,n,a[i]+1,n);
        if(t1>=1&&t1<=n) anss[t1]+=h[i];
        update(1,1,n,1,a[i],i);
    }
    for(int i=1;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=1,ans=0;char c;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){ans=ans*10+c-'0';c=getchar();}
    return f*ans;
}
int c,n;
int x[1001],y[1001],minn=2<<30-1,xp[1001];
struct node{
    int x,y;
}st[1001];
bool CHECK(int l,int r,int w){
    if(r-l+1<c) return 0;
    int cnt=0;
    int yp[1001];
    memset(yp,0,sizeof(yp));
    for(int i=l;i<=r;i++) yp[++yp[0]]=st[i].y;
    sort(yp+1,yp+yp[0]+1);
    for(int i=c;i<=yp[0];i++){
        if(yp[i]-yp[i-c+1]<=w) {
            return 1;
        }
    }
    return 0;
}
bool cmp(node x1,node x2){return x1.x<x2.x;}
bool check(int xx){
    int l=1;
    for(int i=1;i<=n;i++) st[i].x=x[i],st[i].y=y[i];
    sort(st+1,st+n+1,cmp);
    for(int r=1;r<=n;r++){
        if(st[r].x-st[l].x>xx){
            if(CHECK(l,r-1,xx)) {
                return 1;
            }
            while(st[r].x-st[l].x>xx) l++;
        }
    }
    if(CHECK(l,n,xx)) return 1;
    return 0;
}
int main(){
    c=read(),n=read();
    for(int i=1;i<=n;i++) x[i]=read(),y[i]=read();
    int l=0,r=10001;
    while(l<=r){
        int mid=l+r>>1;
        if(check(mid)) minn=min(minn,mid),r=mid-1;
        else l=mid+1;
    }
    cout<<minn+1;
}