优化bellman_ford的方法

bellman_ford(== 边数限制时用 ==):
每次更新每条边，存在负权边的时候dist[n] !=0x3f3f3f3f, 要用 dist[n] > 0x3f3f3f3f / 2来判断
psfa:
优点：万金油，正权边，负权边都可以用，速度快–O mn
缺点：正权边被出题人卡的话，只能用堆优化的dijksktra搞–O mlogn
每次只更新dist[j] > dist[t] + w[i]（距离变短）点所指向的点，若最后dist[n]未被更新则一定为0x3f3f3f3f, 故可以：if(dist[n] == 0x3f3f3f3f) return -1;

C++ 代码

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;

const int N = 1e5 + 10;

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

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

int spfa()
{
    queue<int> q;

    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    q.push(1);
    st[1] = true;

    while(q.size())
    {
        int t = q.front();
        //被弹出后就将其出队
        q.pop();
        st[t] = false;

        //遍历t所指向的每个点的距离
        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];
                //如果该点被更新且没有在队列里面，就将其入队
                if(!st[j])
                {
                    q.push(j);
                    st[j] = true;
                }
            }
        }
    }
    if(dist[n] == 0x3f3f3f3f) return  -1;
    else return dist[n];
}

int main()
{
    memset(h, -1, sizeof(h));

    cin.tie(0);
    cin >> n >> m;

    while(m--)
    {
        cin >> a >> b >> c;
        add(a, b, c);
    }

    int t = spfa();
    if(t == -1) puts("impossible");
    else cout<< t;
}