算法分析：

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Ac代码如下：

这道题是求最长路。采用一个玄学优化：
将新入队节点分为两类，边为正值加到队头，边为负值加到队尾

#pragma GCC optimize(2)
#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <deque>
#include <stack>
using namespace std;

const int N = 100005;
const int K = 200005;
int h[N], ne[K], egs[K], egw[K];
//int ingre[N];
int idx;
int dis[N];
int exst[N];
int cnt[N];
int n, k;


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

long long spfa(){
    deque<int> q;
    for(int i = 1; i <= n; i++){
        dis[i] = 1;
        q.push_back(i);
        exst[i] = 1;
    }

    while(q.size()){
        int u = q.front();
        q.pop_front();
        exst[u] = 0;
        for(int cre = h[u]; cre; cre = ne[cre]){
            int v = egs[cre];
            int w = egw[cre];
            int nw = dis[u] + w;
            if(nw > dis[v]){
                dis[v] = nw;
                cnt[v] = cnt[u]+1;
                if(!exst[v]){
                    if(egw[cre] < 0)q.push_back(v);
                    else q.push_front(v);
                }
                if(cnt[v] > n)
                    return -1;
            }
        }
    }
    long long ans= 0;
    for(int i =1 ; i <= n; i++){
        ans += dis[i];
    }
    return ans;
}

int main(void){
    scanf("%d%d", &n, &k);
//  uno_init();
    for(int i = 0; i < k; i++){
        int x, a, b;
        scanf("%d%d%d", &x,&a, &b);
        switch(x){
            case 1:
                add(a, b, 0);
                add(b, a, 0);
            //  ingre[a]++, intgre[b]++;
                break;
            case 2:
                add(a, b, 1);
            //  add(b, a, 1);
                break;
            case 3:
                add(b, a, 0);
                break;
            case 4:
                add(b, a, 1);
                break;
            case 5:
                add(a, b, 0);
                break;
        }
    }

    printf("%lld", spfa());

}