The first game of 2021 ICPC online game

A Busiest Computing Nodes

The question ：

Yes k Time slice number is 0 To k-1, Yes n A mission , For each task i（ from 0 Numbered starting ）, From i%k After the first time slice starts, find the first free time slice to run , To k-1 The next one is 0.

Line segment tree + Two points

Two points to find the position closest to the left

#include<bits/stdc++.h>
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
using namespace std;
typedef long long ll;
const int maxn=1e5+5;
const int INF=0x3f3f3f3f;
int k,n,h,m=1,mn[maxn<<2],prf[maxn],he[maxn];
void update(int rt,int l,int r,int x,int val){
    
	if(l==r){
    
		mn[rt]=val;
		return;
	}
	int mid=(l+r)>>1;
	if(x<=mid)update(lson,x,val);
	else update(rson,x,val);
	mn[rt]=min(mn[rt<<1],mn[rt<<1|1]);
}
int query(int rt,int l,int r,int x,int y){
    
	if(x<=l&&r<=y)return mn[rt];
	int mid=(l+r)>>1,ans=INF;
	if(x<=mid)ans=min(ans,query(lson,x,y));
	if(y>mid)ans=min(ans,query(rson,x,y));
	return ans;
}
int main(){
    
	scanf("%d%d",&k,&n);
	for(int i=0,x,y;i<n;i++){
    
		scanf("%d%d",&x,&y);
		if(i<k){
    
			update(1,0,k-1,i,x+y);he[i]++;
			continue;
		}
		if(mn[1]>x)continue;
		int d=i%k,l,r,ans;
		int you=query(1,0,k-1,d,k-1);
		if(you<=x)
		    l=d,r=k-1;
		else 
		    l=0,r=d-1; 
		while(l<=r){
    
			int mid=(l+r)>>1;
			if(query(1,0,k-1,l,mid)<=x)ans=mid,r=mid-1;
			else l=mid+1;
		}
		update(1,0,k-1,ans,x+y),he[ans]++,m=max(m,he[ans]);
	}
	for(int i=0;i<k;i++)
	    if(he[i]==m)prf[++h]=i;
	for(int i=1;i<=h;i++){
    
		if(i!=1)printf(" ");
		printf("%d",prf[i]);
	}
}

D. Edge of Taixuan

The question ：

One 1 To n Numbered drawing , For each [L,R] Any two points in the have a weight of $w_i$ The edge of , Now give m Yes $[L_i,R_i]$, After deleting some edges , The remaining edges form the minimum spanning tree .

Ideas ：

Line segment tree + thinking

hold m individual $[L_i,R_i]$ according to $w_i$ Descending order , The back side turns the front side （ Section ） Cover

#include<bits/stdc++.h>
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
using ll=long long;
using namespace std;
const int maxn=1e5+5;
const int INF=0x3f3f3f3f;
int t,n,m,c,mn[maxn<<2],lazy[maxn<<2];
struct node{
    
	int l,r,w;
}e[maxn];
bool cmp(node a,node b){
    
	return a.w>b.w;
}
void pushdown(int rt){
    
	if(lazy[rt]){
    
		mn[rt<<1]=mn[rt];
	    mn[rt<<1|1]=mn[rt];
	    lazy[rt<<1]=lazy[rt];
	    lazy[rt<<1|1]=lazy[rt];
	    lazy[rt]=0;
	}
}
void update(int rt,int l,int r,int x,int y,int val){
    
	if(x<=l&&r<=y){
    
		mn[rt]=val;
		lazy[rt]=val;
		return;
	}
	int mid=(l+r)>>1;
	pushdown(rt);
	if(x<=mid)update(lson,x,y,val);
	if(y>mid)update(rson,x,y,val);
}
int query(int rt,int l,int r,int x){
    
	if(l==r)return mn[rt];
	int mid=(l+r)>>1;
	pushdown(rt);
	if(x<=mid)return query(lson,x);
	else return query(rson,x);
}
int main(){
    
	scanf("%d",&t);
	while(t--){
    
		ll he=0,sum=0,f=0;
		scanf("%d%d",&n,&m);
		memset(mn,INF,sizeof(mn));
		memset(lazy,0,sizeof(lazy));
		for(int i=1;i<=m;i++){
    
			scanf("%d%d%d",&e[i].l,&e[i].r,&e[i].w);
			he+=1ll*e[i].w*(e[i].r-e[i].l+1)*(e[i].r-e[i].l)/2;
		}
		sort(e+1,e+1+m,cmp);
		for(int i=1;i<=m;i++)
		    update(1,1,n-1,e[i].l,e[i].r-1,e[i].w);
		for(int i=1;i<n;i++){
    
			int h=query(1,1,n-1,i);
			if(h==INF)f=1;
			sum+=h;
		}
		printf("Case #%d: ",++c);
		if(f)printf("Gotta prepare a lesson");
		else printf("%lld",he-sum);
		if(t)printf("\n");
	}
}

H. Mesh Analysis

Handwriting discretization

#include<bits/stdc++.h>
using namespace std;
const int maxn=2e6+5;
struct node{
    
	int id,code,x,y,z;
}e[maxn];
int n,m,t,k,p[maxn],a[maxn],b[maxn];
vector<int>gap[maxn];
map<int,int>mp1,mp2;
set<int>s1,s2;
int main(){
    
	scanf("%d%d",&n,&m);
	for(int i=1,s;i<=n;i++){
    
		double x,y,z;
		scanf("%d%lf%lf%lf",&s,&x,&y,&z);
	}
	for(int i=1,id,code,x,y,z;i<=m;i++){
    
		scanf("%d%d%d%d",&id,&code,&x,&y);
		e[i]={
    id,code,x,y,0};
		a[++k]=id,a[++k]=x,a[++k]=y;
		if(code==203){
    
			scanf("%d",&z);
			e[i].z=z;
			a[++k]=z;
		}
	}
	scanf("%d",&t);
	for(int i=1;i<=t;i++){
    
		scanf("%d",&p[i]);
		a[++k]=p[i];
	}
	for(int i=1;i<=k;i++)b[i]=a[i];
	sort(b+1,b+1+k);
	int d=1;
	for(int i=2;i<=k;i++){
    
		if(b[i]!=b[i-1])b[++d]=b[i];
	}
	for(int i=1;i<=k;i++){
    
		int s=lower_bound(b+1,b+1+d,a[i])-b;
		mp1[a[i]]=s;
		mp2[s]=a[i];
	}
	for(int i=1;i<=m;i++){
    
		int a=mp1[e[i].x],b=mp1[e[i].y],c=mp1[e[i].z];
		gap[a].push_back(b);
		gap[b].push_back(a);
		if(e[i].code==203){
    
			gap[a].push_back(c);
			gap[b].push_back(c);
			gap[c].push_back(a);
			gap[c].push_back(b);
		}
	}
	for(int i=1;i<=t;i++){
    
		s1.clear(),s2.clear();
		int q=p[i],dui=mp1[q];
		printf("%d\n",q);
		if(dui==0){
    
			printf("[][]");
		}
		else{
    
			for(int i=0;i<gap[dui].size();i++){
    
				s1.insert(mp2[gap[dui][i]]);
			}
			int f=0;
			printf("[");
			for(auto x:s1){
    
				if(f)printf(",");
				printf("%d",x);
				f=1;
			}
			printf("]\n");
			for(int j=1;j<=m;j++){
    
				if(e[j].x==p[i]||e[j].y==p[i]||e[j].z==p[i])s2.insert(e[j].id);
			}
			f=0;
			printf("[");
			for(auto x:s2){
    
				if(f)printf(",");
				printf("%d",x);
				f=1;
			}
			printf("]");
		}
		if(i!=t)printf("\n");
	}
}

map+set

map<int,set<int>>mp;

mp[123].insert()

for(auto &x:mp[123]){
    
}