acwing 853. 有边数限制的最短路 模板

输入样例：
3 3 1
1 2 1
2 3 1
1 3 3
输出样例：
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;

    //进行n轮
    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 ret = solve();

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


    return 0;
}