Dijkstra求最短路—priority堆优化；

输入样例：

3 3
1 2 2
2 3 1
1 3 4

输出样例：

3

 

#include
using namespace std;
const int N = 3e5, inf = 0x3f3f3f3f;
typedef pair<int, int> pii;
int n, m, e[N], ne[N], h[N], w[N], dis[N],idx;
bool st[N];
 
void add(int a, int b, int c) {
	w[idx] = c;
	e[idx] = b;
	ne[idx] = h[a];
	h[a] = idx++;
}
 
int dijkstra() {
	memset(dis, 0x3f, sizeof dis);
	priority_queue, greater >heap;
	heap.push({ 0,1 });
	dis[1] = 0;//初始化起点位置；
	while (heap.size()) {
		auto now = heap.top();
		heap.pop();
		int pos = now.second, dist = now.first;
		if (st[pos]) continue;
		st[pos] = true;
		for (int i = h[pos];i != -1;i = ne[i]) {
			int j = e[i];
			if (dis[j] > dist + w[i]) {//w[i]表示到i点这条边的距离而不是w[j]
				dis[j] = dist + w[i];	//i只是个下标，e中在存的是i这个下标对应的点
				heap.push({ dis[j],j });
			}
		}
	}
 
	if (dis[n] == inf) return -1;
	else return dis[n];
}
 
 
 
int main() {
	memset(h, -1, sizeof h);
	cin >> n >> m;
	for (int i = 0;i < m;i++) {
		int a, b, c;
		cin >> a >> b >> c;
		add(a, b, c);
	}
	cout << dijkstra() << endl;
}