luogu 1484\1792 种树 奇怪的贪心可反悔

1484 种树 此版本是线性的，那么根据链表维护即可；

构建新点，点的左右分别是原整个区间的前驱及后继，再正常维护即可

注意两个版本的维护有所不同

第二个版本的维护直接将左右两点删除

1792 种树2  此版本是环

1484

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=;
struct node{
    ll v;int id;
    bool operator <(node a)const{
    return v<a.v;}};
inline ll read(){
    ll x=,f=;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-;ch=getchar();}
    while(isdigit(ch)){x=(x<<)+(x<<)+(ch^);ch=getchar();}
    return x*f;}
ll a[N],ans;
int l[N],r[N],vis[N];
priority_queue<node> q;
int n,k;
int main(){
    n=read();k=read();
    for(int i=;i<=n;i++){
        a[i]=read();
        q.push((node){a[i],i});
        l[i]=i-;r[i]=i+;
    }int len=n;
    for(int i=;i<=k;i++){
        while(!q.empty()&&vis[q.top().id]) q.pop();
        if(q.empty()||q.top().v<)
            break;
        node u=q.top();q.pop();
        ans+=u.v;
        vis[u.id]=vis[l[u.id]]=vis[r[u.id]]=;
        a[++len]=a[l[u.id]]+a[r[u.id]]-a[u.id];
        l[len]=l[l[u.id]],r[len]=r[r[u.id]];
        r[l[len]]=len;l[r[len]]=len;
        q.push((node){a[len],len});
    }printf("%lld\n",ans);
    return ;
}

1792

#include<bits/stdc++.h>
#define M 200050
#define rep(i,x,y) for(register int i=x;i<=y;i++)
using namespace std;
inline int read(){
    int x=,f=;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-;ch=getchar();}
    while(isdigit(ch)){x=(x<<)+(x<<)+(ch^);ch=getchar();}
    return x*f;}
struct node{
    int v,id;
    friend bool operator < (node x,node y){
        return x.v < y.v;}
};priority_queue<node> q;
int n,m,ans,chose,r[M],l[M],a[M];
bool vis[M];
void change(int x){
    vis[x]=;
    r[l[x]]=r[x];
    l[r[x]]=l[x];
    r[x]=;l[x]=;}
int main(){
    n=read(),m=read();
    rep(i,,n) a[i]=read();
    if((n>>)<m){
        printf("Error!\n");return ;}
    rep(i,,n-) r[i]=i+;r[n]=;
    rep(i,,n) l[i]=i-;l[]=n;
    rep(i,,n) q.push((node){a[i],i});
    rep(i,,m){
        while(vis[q.top().id]) q.pop() ;
        node u=q.top();q.pop();
        ans+=u.v;
        int ll=l[u.id],rr=r[u.id];
        a[u.id]=a[ll]+a[rr]-a[u.id];
        change(ll);change(rr);
        q.push((node){a[u.id],u.id});
    }printf("%d\n",ans);return ;
}