acwing 853. There shortest limit the number of sides of the template

Address https://www.acwing.com/problem/content/description/855/

Given a point n and m edges directed graph, there may be multiple edges drawing and loopback, the edge weights may be negative.

After you find the shortest distance at most k edges, and if you can not go from point n No. 1 point, 1-point output from the impossible to the number n points.

Note: there may be negative cycles figure.

Input Format

The first line contains three integers n, m, k.

Subsequently m rows, each row containing three integers x, y, Z, it indicates the presence of a point x to point y directed edge from a side length of z.

Output Format

Output an integer, k represents the shortest distance through the up edges from point 1 to point number n.

If the path does not satisfy the condition, it outputs "impossible".

data range

Sample input:
 . 3  . 3  . 1 
. 1  2  . 1 
2  . 3  . 1 
. 1  . 3  . 3
Sample output:
3

#include <iostream>
#include <vector>
#include <memory.h>


using namespace std;

const  int N = 510 ;

vector<vector<pair<int, int>>> v;
int dist[N];
int back[N];
int n, m, k;

/*
3 3 1
1 2 1
2 3 1
1 3 3
*/
void solveInner(int back[])
{
    for (int i = 0; i < v.size(); i++) {
        for (int j = 0; j < v[i].size(); j++) {
            int a = i; int b = v[i][j].first; int w = v[i][j].second;
            if (dist[b] > back[a] + w) {
                dist[b] = back[a] + w;
            }
        }
    }
}


int solve()
{
    memset(dist, 0x3f, sizeof(dist));
    dist[1] = 0;

    // for n-th wheel 
    for ( int I = 0 ; I <K; I ++ ) {
        memcpy(back, dist, sizeof dist);
        solveInner (back);
    }

    if (dist[n] > 0x3f3f3f3f / 2) return -1;
    return dist[n];

}


int main ()
{
    scanf("%d%d%d",&n,&m,&k);
    //cin >> n >> m >> k;

    v.resize(n + 1);

    for (int i = 0; i < m; i++) {
        int a, b, c;
        //cin >> a >> b >> c;
        scanf("%d%d%d",&a,&b,&c);
        v[a].push_back({ b,c });
    }

    int K = solve ();

    if (ret == -1) printf("impossible");
    else printf("%d\n",ret); 


    return 0;
}