csp 201912-4区块链 80分

思路

直接利用优先队列，将每一次会对外传播的节点当前的状态记住，同时也存储这次传播会到达的时间
每次产生和输出之前先更新就可以,但是因为一个节点可能在同一时间生成多个块，会被卡时间，只能得到80分，后续优化懒得做了
附赠一组测试数据

---

15 13
1 2
2 3
3 4
4 5
1 6
6 7
7 8
8 9
1 10
10 11
11 12
12 13
14 15
6 28
1 1 1
1 2 2
1 6
2 7
13 7
9 7
5 7
3 14
8 14
5 14
11 14
9 25
5 25
13 25
9 29 3
5 29 4
13 29 5
1 53
2 59 6
2 59
1 1000
3 1000
8 1000
9 1000
10 1000
13 1000
14 1000
15 1000

---

输出

3 0 1 2
2 0 1
1 0
1 0
1 0
3 0 1 2
1 0
1 0
3 0 1 2
2 0 1
2 0 1
2 0 1
4 0 1 2 3
5 0 1 2 3 6
5 0 1 2 3 6
5 0 1 2 3 6
5 0 1 2 3 6
5 0 1 2 3 6
5 0 1 2 3 6
5 0 1 2 3 6
1 0
1 0

---

代码

#include<bits/stdc++.h>
using namespace std;
struct event_bro
{
	int id;
	int t;
	vector<int> gg;
	event_bro(int id,int t):id(id),t(t){
	}
	bool operator<(const event_bro&a) const
	{
		return t>a.t;
	}
};
vector<int> temp[505];
vector<int> Graph[505];
priority_queue<event_bro> pq;
void handler(int t,int gap){ 
	while(!pq.empty()&&pq.top().t<=t){
		event_bro he = pq.top();
		pq.pop();
		for(auto &i:Graph[he.id]){
			if(temp[i].size()<he.gg.size()||
				temp[i].size()==he.gg.size()&&
				temp[i][temp[i].size()-1]>he.gg[he.gg.size()-1]){
				temp[i]=he.gg;
				event_bro __(0,0);
				__.id=i;
				__.t=he.t+gap;
				__.gg = he.gg;
				pq.push(__);
			}
		}
	}
}
void query(int a,int b,int t){
	handler(b,t);
	cout<<temp[a].size()<<" ";
	for(auto &i:temp[a])
		cout<<i<<" ";
	cout<<endl;
}
void update(int a,int b,int c,int t){
	handler(b,t);
	temp[a].push_back(c);
	event_bro help(a,b+t);
	help.gg=temp[a]; 
	pq.push(help);
}
int main(int argc, char const *argv[])
{
	freopen("1.dat","r",stdin);
	freopen("ans.dat","w",stdout);
	int m,n;
	cin>>m>>n;
	int a,b,c;
	for (int i = 0; i < 502; ++i)
		temp[i].push_back(0);
	for (int i = 0; i < n; ++i)
	{
		cin>>a>>b;
		Graph[a].push_back(b);
		Graph[b].push_back(a);
	}
	int t,k;
	cin>>t>>k;
	while(k--){
		cin>>a>>b;
		// char ______=getchar();
		if(getchar()=='\n'||cin.eof()){
			query(a,b,t);
		}else{
			cin>>c;
			update(a,b,c,t);
		}
	}
	return 0;
}