Title Link: (bzoj) https://www.lydsy.com/JudgeOnline/problem.php?id=4386
(luogu) https://www.luogu.org/problemnew/show/P3597

Why can't I do this?
First of all, if all the edge weights are \ (1 \), a new counter will be created, with each point connected to the counter and the counter connected to the self loop. Then the adjacent matrix can be multiplied by a fast power
If the edge has \ (2 \) and \ (3 \), create a new node to connect to the outgoing point
There are many details, especially when judging whether it is greater than \ (k \), long long cannot be exploded (it is said that this data water, so I can't guarantee that the following code won't be stuck)

Code

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cassert>
#include<iostream>
#define llong long long
using namespace std;

inline int read()
{
    int x=0; bool f=1; char c=getchar();
    for(;!isdigit(c);c=getchar()) if(c=='-') f=0;
    for(; isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+(c^'0');
    if(f) return x;
    return -x;
}

const int N = 121;
llong p;
struct Matrix
{
    llong a[N+3][N+3]; int n;
    Matrix() {}
    Matrix(int _n) {n = _n; for(int i=0; i<=n; i++) for(int j=0; j<=n; j++) a[i][j] = 0ll;}
    void init(int _n) {n = _n; for(int i=0; i<=n; i++) for(int j=0; j<=n; j++) a[i][j] = 0ll;}
    void unitize(int _n) {n = _n; a[0][0] = 0ll; for(int i=1; i<=n; i++) for(int j=1; j<=n; j++) a[i][j] = i==j?1ll:0ll;}
    void output() {if(a[0][0]==-1) puts("gg"); for(int i=1; i<=n; i++) {for(int j=1; j<=n; j++) printf("%d ",a[i][j]); puts("");}}
    Matrix operator *(const Matrix &arg) const
    {
        Matrix ret(n);
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
            {
                for(int k=1; k<=n; k++)
                {
                    if(k==n)
                    {
                        if(arg.a[j][k]>0 && (a[i][j]>=p/arg.a[j][k]+1 || ret.a[i][k]>=p-a[i][j]*arg.a[j][k]))
                        {
                            ret.a[0][0] = -1;
                            return ret;
                        }
                    }
                    ret.a[i][k] = ret.a[i][k]+a[i][j]*arg.a[j][k];
                }
            }
        }
        llong sum = 0ll;
        for(int i=1; i<=n/3; i++)
        {
            if(sum>=p-ret.a[i][n]) {ret.a[0][0] = -1; return ret;}
            sum += ret.a[i][n];
        }
        return ret;
    }
} g,pw[65],tmp;
int n,m;

llong solve()
{
    pw[0] = g; int i = 0;
    for(i=1; i<=61; i++)
    {
        pw[i] = pw[i-1]*pw[i-1];
        if(pw[i].a[0][0]==-1)
        {
            break;
        }
    }
    if(i==62) {return -1;}
    llong ret = 0ll; g.unitize(n+n+n+1);
    for(i--; i>=0; i--)
    {
        tmp = g*pw[i];
        if(tmp.a[0][0]!=-1)
        {
            ret|=(1ll<<i);
            g = tmp;
        }
    }
    return ret;
}

int main()
{
    scanf("%d%d%lld",&n,&m,&p); p+=n;
    g.init(n+n+n+1);
    for(int i=1; i<=m; i++)
    {
        int x,y,w; scanf("%d%d%d",&x,&y,&w);
        if(w==1) {g.a[x][y]++;}
        else if(w==2) {g.a[x+n][y]++;}
        else if(w==3) {g.a[x+n+n][y]++;}
    }
    for(int i=1; i<=n; i++)
    {
        g.a[i][i+n]++;
        g.a[i+n][i+n+n]++;
        g.a[i][n+n+n+1]++;
    }
    g.a[n+n+n+1][n+n+n+1]++;
    llong ans = solve();
    printf("%lld\n",ans);
    return 0;
}