分层图的四倍经验

做完冻结后在做这三道题，简直爆切，直接四倍经验(STO)。

飞行路线：

几乎跟冻结一模一样就不讲啦

#include <bits/stdc++.h>
using namespace std;
int n , m , k , ans = 0x3ffffff , s , ee;
int dis[10000010] , vis[10000010];
vector<pair<int , int> > e[1000010];
void work(){
	priority_queue<pair<int , int> > q; 
	memset(dis , 127 , sizeof(dis));
	dis[s + k * n] = 0;
	q.push(make_pair(0 , s + k * n));
	while(!q.empty()){
		int x = q.top().second;
		q.pop();
		if(vis[x]) continue;
		vis[x] = 1;
		for(int i = 0; i < e[x].size(); i++){
			int nx = e[x][i].first , w = e[x][i].second;
			if(dis[nx] > dis[x] + w){
				dis[nx] = dis[x] + w;
				q.push(make_pair(-dis[nx] , nx));
			}
		}
	}
}
int main(){
	cin >> n >> m >> k >> s >> ee;
	s++ , ee++;
	for(int i = 1; i <= m; i++){
		int x , y , z;
		cin >> x >> y >> z;
		x++ , y++;
		e[x].push_back(make_pair(y , z));
		e[y].push_back(make_pair(x , z));
		for(int j = 1; j <= k; j++){
			e[x + j * n].push_back(make_pair(y + n * (j - 1) , 0));
			e[x + j * n].push_back(make_pair(y + n * j , z));
			e[y + j * n].push_back(make_pair(x + n * (j - 1) , 0));
			e[y + j * n].push_back(make_pair(x + n * j , z));
		}
	}
	work();
	for(int i = 0; i <= k; i++) ans = min(dis[ee + i * n] , ans);
	cout << ans;
	return 0;
}

Telephone Lines S：

这道题改了一点点，就是让你输出的是最长的一条电话线，在跑最短路的时候改一下就OK了：

#include <bits/stdc++.h>
using namespace std;
int n , m , k , ans = 0x3ffffff , s , ee;
int dis[10000010] , vis[10000010];
vector<pair<int , int> > e[1000010];
void work(){
	priority_queue<pair<int , int> > q; 
	memset(dis , 127 , sizeof(dis));
	dis[s + k * n] = 0;
	q.push(make_pair(0 , s + k * n));
	while(!q.empty()){
		int x = q.top().second;
		q.pop();
		if(vis[x]) continue;
		vis[x] = 1;
		for(int i = 0; i < e[x].size(); i++){
			int nx = e[x][i].first , w = e[x][i].second;
			if(dis[nx] > dis[x] + w){
				dis[nx] = dis[x] + w;
				q.push(make_pair(-dis[nx] , nx));
			}
		}
	}
}
int main(){
	cin >> n >> m >> k >> s >> ee;
	s++ , ee++;
	for(int i = 1; i <= m; i++){
		int x , y , z;
		cin >> x >> y >> z;
		x++ , y++;
		e[x].push_back(make_pair(y , z));
		e[y].push_back(make_pair(x , z));
		for(int j = 1; j <= k; j++){
			e[x + j * n].push_back(make_pair(y + n * (j - 1) , 0));
			e[x + j * n].push_back(make_pair(y + n * j , z));
			e[y + j * n].push_back(make_pair(x + n * (j - 1) , 0));
			e[y + j * n].push_back(make_pair(x + n * j , z));
		}
	}
	work();
	for(int i = 0; i <= k; i++) ans = min(dis[ee + i * n] , ans);
	cout << ans;
	return 0;
}

Revamping Trails G：

也是很简单的，我就不水了：

#include <bits/stdc++.h>
using namespace std;
int n , m , k , ans = 0x3ffffff;
int dis[1000010] , vis[1000010];
vector<pair<int , int> > e[1000010];
void work1(){
	priority_queue<pair<int , int> > q; 
	memset(dis , 127 , sizeof(dis));
	dis[1 + k * n] = 0;
	q.push(make_pair(0 , 1 + k * n));
	while(!q.empty()){
		int x = q.top().second;
		q.pop();
		if(vis[x]) continue;
		vis[x] = 1;
		for(int i = 0; i < e[x].size(); i++){
			int nx = e[x][i].first , w = e[x][i].second;
			if(dis[nx] > dis[x] + w){
				dis[nx] = dis[x] + w;
				q.push(make_pair(-dis[nx] , nx));
			}
		}
	}
}
int main(){
	cin >> n >> m >> k;
	for(int i = 1; i <= m; i++){
		int x , y , z;
		cin >> x >> y >> z;
		e[x].push_back(make_pair(y , z));
		e[y].push_back(make_pair(x , z));
		for(int j = 1; j <= k; j++){
			e[x + j * n].push_back(make_pair(y + n * (j - 1) , 0));
			e[x + j * n].push_back(make_pair(y + n * j , z));
			e[y + j * n].push_back(make_pair(x + n * (j - 1) , 0));
			e[y + j * n].push_back(make_pair(x + n * j , z));
		}
	}
	work1();
	for(int i = 0; i <= k; i++) ans = min(dis[n + i * n] , ans);
	cout << ans;
	return 0;
}