题目链接：Dijkstra求最短路 II

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>

using namespace std;

typedef pair<int, int> PII;

const int N = 150010;

int n, m;
int h[N], w[N], e[N], ne[N], idx;

int dist[N];
bool st[N];

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

int dijkstra()
{
    memset(dist, 0x3f, sizeof dist);
    
    dist[1] = 0;
    
    priority_queue<PII, vector<PII>, greater<PII>> heap;
    
    // {0, 1} 距离为0 编号为1
    heap.push({0, 1});
    
    while(heap.size())
    {
        // 小根堆的堆顶元素就是最小元素(根据距离排列)
        auto t = heap.top();
        heap.pop();
        
        int distance = t.first, ver = t.second;
        
        if(!st[ver])
        {
            st[ver] = true;
            // 更新其他点
            for(int i = h[ver]; i != -1; i = ne[i])
            {
                int j = e[i];
                if(dist[j] > distance + w[i])
                {
                    dist[j] = distance + w[i];
                    heap.push({dist[j], j});
                }
            }
        }
    }
    if(dist[n] == 0x3f3f3f3f) return -1;
    return dist[n];
}

int main()
{
    cin >> n >> m;
    
    memset(h, -1, sizeof h);
    
    for(int i = 0; i < m; i ++)
    {
        int a, b, c;
        cin >> a >> b >>c;
        add(a, b, c);
    }
    
    cout << dijkstra() << endl;
    
    return 0;
}