Several greedy Summary

1. Press the end of small to large order: Select the interval, the interval selected point

Examples (the interval selected point) Valley 1250 Los trees

#include<iostream>
#include<cstdio>
using namespace std;
struct strage{
    int kw;
    int jw;
    int t;
}a[30001],temp;
int main(){
    int n,h,i,j,k,sum=0;
    int za[30001]={0};
    scanf("%d%d",&n,&h);
    for(i=1;i<=h;i++)
      scanf("%d%d%d",&a[i].kw,&a[i].jw,&a[i].t);
      
    for(i=1;i<h;i++)
      for(j=i+1;j<=h;j++)
      if(a[i].jw>a[j].jw){temp=a[i];a[i]=a[j];a[j]=temp;}
      
    for(i=1;i<=h;i++)
    {
      for(j=a[i].kw;j<=a[i].jw;j++)
        if(za[j]==1) a[i].t--;
      k=a[i].t;
      if(k>0){
      for(j=a[i].jw;j>=a[i].kw;j--)
        if(za[j]==0)
         {
            za[j]=1;
            k--;
            if(k==0)break;
         }
       }
    }
    
    for(i=1;i<=n;i++)if(za[i]==1)sum++;
    printf("%d",sum);
    return 0;
} 

2. beginning from small to large: interval coverage

3. You can go back on the greedy

Method: 1. build a simulation doubly linked list (a before and after a record)

2. When a point removed from the priority queue, mark about two visited update doubly linked list, right and left + right point right point - the current point right into the team

Example P1792

// luogu-judger-enable-o2
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#define N 200007
using namespace std;
struct tu
{
    int val,l,r;

}p[N];
struct node
{
    int val,pos;
      bool operator<(const node &x) const
    {
        return val<x.val;
    } 
};
int n,m,ans;
bool vis[N];
priority_queue<node> q;
void dui(int x)
{
    p[x].l=p[p[x].l].l;
    p[x].r=p[p[x].r].r;
    p[p[x].l].r=x;
    p[p[x].r].l=x;
}
int main()
{
    scanf("%d%d",&n,&m);
    if(n<m*2)
    {
        printf("Error!");
        return 0;
    }
    for(int i=1;i<=n;++i)
    {
        scanf("%d",&p[i].val);
        p[i].l=i-1;
        p[i].r=i+1;
        q.push((node){p[i].val,i});
    }
    p[1].l=n,p[n].r=1;
    for(int i=1;i<=m;++i)
    {
        while(vis[q.top().pos])
            q.pop();
        node now=q.top();
        q.pop();
        ans+=now.val;
        vis[p[now.pos].l]=vis[p[now.pos].r]=1;
        p[now.pos].val=p[p[now.pos].l].val+p[p[now.pos].r].val-p[now.pos].val;
        q.push((node){p[now.pos].val,now.pos});
        dui(now.pos);
    }

    printf("%d",ans);
    return 0;
}

Or to push it greedy strategy, more than just a few of the classic questions