851. spfa求最短路 题解

#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
const int N = 100010;

int n, m, a, b, c, idx;
int h[N], ne[N], e[N], w[N], dist[N];
bool vis[N];

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

int spfa()
{
    memset (dist, 0x3f, sizeof dist);
    dist[1] = 0;
    
    queue<int>q;
    q.push(1);
    vis[1] = true;
    
    while (q.size())
    {
        int t = q.front();
        q.pop();
        vis[t] = false;  // 从队列中取出来之后该节点st被标记为false,代表之后该节点如果发生更新可再次入队
        
        for (int i = h[t]; i != -1; i = ne[i])
        {
            int j = e[i];
            if (dist[j] > dist[t] + w[i])
            {
                dist[j] = dist[t] + w[i];
                if (!vis[j])  // 当前已经加入队列的结点，无需再次加入队列，即便发生了更新也只用更新数值即可，重复添加降低效率
                {
                    q.push(j);
                    vis[j] = true;    
                }
            }
        }
    }
    if (dist[n] == 0x3f3f3f3f)  return 0;
    return dist[n];
}

int main()
{
    memset (h, -1, sizeof h);
    cin >> n >> m;
    
    while (m -- )
    {
        cin >> a >> b >> c;
        add(a, b, c);
    }
    
    int t = spfa();
    if (t == 0)    puts("impossible");
    else cout << t;
    return 0;
}