AcWing 340. Solution to communication line problem (binary + double ended queue BFS for the shortest circuit)

AcWing 340. Communication lines
y Always say yes noip Improve the level of the Group , It's really difficult to change the meaning of the question , It's hard to find a solution , I hope I can write this level of questions by myself one day
Their thinking ： Two points + deque BFS Find the shortest path , It is difficult to change thinking , Several turning points of thinking are ：① Find a suitable less than k Value x, Two points .② In judging x When appropriate , You can use a double ended queue to find the shortest path

#include<bits/stdc++.h>

using namespace std;

const int N = 1e3 + 10, M = 2e4 + 10, INF = 0x3f3f3f3f;

int h[N], ne[M], e[M], w[M], idx;
bool st[N];
deque<int>q;
int n, m, p, k;
int dist[N];

void add(int a, int b, int c){
    
	e[idx] = b;
	w[idx] = c;
	ne[idx] = h[a];
	h[a] = idx ++ ;
}

bool check(int bound){
      // The weight of the opposite side is only 0/1 The graph uses a double ended queue to search widely to find the shortest path  
	memset(st, 0, sizeof st);
	memset(dist, 0x3f, sizeof dist);
	
	dist[1] = 0;
	q.push_front(1);
	
	while(q.size()){
    
		int op = q.front();
		q.pop_front();
		
		if(st[op]) continue;
		st[op] = true;
		
		for(int i = h[op]; ~i; i = ne[i]){
    
			int j = e[i], v = w[i] > bound;
			if(dist[j] > dist[op] + v){
    
				dist[j] = dist[op] + v;
				if(!v) q.push_front(j);
				else q.push_back(j);
			} 
		}
	}
	return dist[n] <= k;  // The edge weight in the return graph is greater than bound Whether the number of edges of is greater than k 
}

int main()
{
    
	cin>>n>>m>>k;
	memset(h, -1, sizeof h);
	while(m -- ){
    
		int a, b, c;
		cin>>a>>b>>c;
		add(a, b, c);
		add(b, a, c);
	}
	
	// Two points to find the right value mid 
	// The right value mid It means having mid The weight of the edge is assigned 0 when , The shortest path can be found 
	// Here I have always wondered why the dichotomy mid It must be the weight in the graph 
	// After reading the detailed explanation of other bosses, I understand 
	//① The first understanding , If mid It is not the weight existing in the graph , So this time check The result is the same as last check The results returned are the same , Two points won't stop 
	//② The second understanding , The essence of dichotomy is to find the minimum , If mid Not the edge weights that exist in the graph , Then there must be a smaller number of edge weights , Chapter II continued , Until the smallest qualified value is found 
	// In a word , Two points is the best value , Instead of being appropriate  
	int l = 0, r = 1e6 + 1;
	while(l < r){
    
		int mid = l + r >> 1;
		if(check(mid)) r = mid;
		else l = mid + 1;
	}
	
	if(r == 1e6 + 1) r = -1;
	cout<<r<<endl;
	
	return 0;
}