I Incoming Asteroids(优先队列)

Posted 2021-12-20 Harris-H

I Incoming Asteroids(优先队列)

考虑递增的速度，用一个定时器维护每个camera的观测值，当到达阈值之后，就进行检测，每个位置用一个优先队列维护。

// Problem: I. Incoming Asteroids
// Contest: Codeforces - 2019-2020 ICPC Asia Hong Kong Regional Contest
// URL: https://codeforces.ml/gym/102452/problem/I
// Memory Limit: 512 MB
// Time Limit: 2000 ms
// Date: 2021-11-25 22:17:28
// --------by Herio--------

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull; 
const int N=2e5+5,M=2e4+5,inf=0x3f3f3f3f,mod=1e9+7;
const int hashmod[4] = 402653189,805306457,1610612741,998244353;
#define mst(a,b) memset(a,b,sizeof a)
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define x first
#define y second
#define pb emplace_back
#define SZ(a) (int)a.size()
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define ios ios::sync_with_stdio(false),cin.tie(nullptr) 
void Print(int *a,int n)
	for(int i=1;i<n;i++)
		printf("%d ",a[i]);
	printf("%d\\n",a[n]); 

template <typename T>		//x=max(x,y)  x=min(x,y)
void cmx(T &x,T y)
	if(x<y) x=y;

template <typename T>
void cmn(T &x,T y)
	if(x>y) x=y;

int n,m,last;
int a[N][4],cnt;
int goal[N];
bool vis[N];
ll tim[N];
priority_queue<PII,vector<PLL>,greater<PLL> >q[N];
void solve(int x)
	vector<int>ans;
	while(!q[x].empty()&&tim[x]>=q[x].top().x)
		auto u=q[x].top();q[x].pop();
		int id = u.y;
		if(vis[id]) continue;
		ll rest = goal[id] ;
		rep(i,1,a[id][0]) rest -= tim[a[id][i]];
		if(rest<=0)
			ans.pb(id);
			vis[id]=1;
			continue;
		
		ll delta = (rest+a[id][0]-1)/a[id][0];
		rep(i,1,a[id][0]) q[a[id][i]].push((tim[a[id][i]]+delta),id);
	
	printf("%d",SZ(ans));last=SZ(ans);sort(ans.begin(),ans.end());
	for(int x:ans) printf(" %d",x);
	puts("");

int main()
	scanf("%d%d",&n,&m);
	while(m--)
		int op;scanf("%d",&op);
		if(op==1)
			int y,k;scanf("%d%d",&y,&k);
			y^=last;
			goal[++cnt]=y;
			a[cnt][0]=k;
			ll delta = (y+k-1)/k;
			rep(j,1,k)
				scanf("%d",&a[cnt][j]);
				a[cnt][j]^=last;
				goal[cnt]+=tim[a[cnt][j]];//这里要加上之前的观测值
                //因为是从现在开始算,前面的不算贡献
			
			rep(j,1,k)
				q[a[cnt][j]].push((tim[a[cnt][j]]+delta),cnt);
			
			
		else 
			int x,y;scanf("%d%d",&x,&y);
			x^=last,y^=last;
			tim[x]+=y;
			solve(x);
		
	
	return 0;