P2672 推销员

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;

struct home{
    int s,v;
}a[100010];
int q[100010];
int h[100010],qm[100010];
int n;

bool cmp(home a,home b){
    return a.v>b.v;
}

int main(){
    scanf("%d",&n);
    for (int i=1;i<=n;i++){
        scanf("%d",&a[i].s);
    }
    for (int i=1;i<=n;i++){
        scanf("%d",&a[i].v);
    }
    sort(a+1,a+1+n,cmp);
    for (int i=n;i>=1;i--){
        h[i]=max(h[i+1],2*a[i].s+a[i].v);
    }
    for (int i=1;i<=n;i++){
        qm[i]=max(qm[i-1],a[i].s);
    }
    for (int i=1;i<=n;i++){
        q[i]=q[i-1]+a[i].v;
    }
    for (int i=1;i<=n;i++){
        printf("%d\n",max(q[i-1]+h[i],q[i]+2*qm[i]));
    }
} 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<algorithm>
using namespace std;
const int N=100005;
const int INF=599518803;

int n;
int s[N],a[N];
struct Tree
{
    int l,r,mid,lazy;
    int id,val;
}tree[N<<2];
int last,now;

int read(){//快读
    char c=getchar();int num=0;
    for(;!isdigit(c);c=getchar());
    for(;isdigit(c);c=getchar())
        num=num*10+c-‘0‘;
    return num;
}

void push_up(int root){
    if(tree[root<<1].val>=tree[root<<1|1].val){
        tree[root].id=tree[root<<1].id;
        tree[root].val=tree[root<<1].val;
    }
    else{
        tree[root].id=tree[root<<1|1].id;
        tree[root].val=tree[root<<1|1].val;
    }
}

void build(int root,int l,int r){
    tree[root].l=l,tree[root].r=r,tree[root].mid=l+r>>1;
    if(l==r){
        tree[root].id=l;
        tree[root].val=(s[l]<<1)+a[l];
        return;
    }
    build(root<<1,l,tree[root].mid);
    build(root<<1|1,tree[root].mid+1,r);
    push_up(root);
}

void push_down(int root){
    tree[root<<1].lazy+=tree[root].lazy;
    tree[root<<1|1].lazy+=tree[root].lazy;
    tree[root<<1].val+=tree[root].lazy;
    tree[root<<1|1].val+=tree[root].lazy;
    tree[root].lazy=0;
    return;
}

void update(int root){
    if(tree[root].r<=last)    //因为在last之前的一定已经被走过了，已经被置为了a[i]，没必要再次置一次，所以直接返回 
        return;
    if(tree[root].l>now){    //在当前的人右边的 
        tree[root].val-=(s[now]-s[last])<<1;    //减路程 
        tree[root].lazy-=(s[now]-s[last])<<1;    //lazy数组也要减 
        return;
    }
    if(tree[root].l==tree[root].r){        //在last和now之间的，换成他们的推销值
        tree[root].val=a[tree[root].l];
        return;
    }
    if(tree[root].lazy){
        push_down(root);
    }
    update(root<<1);
    update(root<<1|1);
    push_up(root);
}
void _delete(int root){    //将now删除 
    if(tree[root].l==tree[root].r){
        tree[root].val=-INF;    //val置为极小值 
        return;
    }
    if(now<=tree[root].mid)
        _delete(root<<1);
    else
        _delete(root<<1|1);
    push_up(root);
}

int main(){
    n=read();
    for(int i=1;i<=n;++i)
        s[i]=read();
    for(int i=1;i<=n;++i)
        a[i]=read();
    build(1,1,n);
    int ans=0;
    for(int i=1;i<=n;++i){
        ans+=tree[1].val;    //取出最大的贡献 
        now=tree[1].id;
        printf("%d\n",ans);
        if(now>last){
            update(1);
            last=now;    //更新last 
        }
        _delete(1);    //删除now 
    }
    return 0;
}