AcWing 853. 有边数限制的最短路 bellman-ford

//存在负权值   处理负环 
//如果能求出来  一般是不存在负权回路
//如果有负回路  那最小距离可能是负无穷
#include <cstring>
#include <iostream>
using namespace std;
const int N = 1e4+1;
struct Edge {
    int a;
    int b;
    int w;
} edge[N];
int n, m, k;
int dist[N], backup[N];
void bellman_ford() {
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    ////迭代k次，表示经过不超过k条边走到每个点的距离
    for (int i=0; i<k; i++) {
        memcpy(backup , dist, sizeof dist);//备份，不加备份可能出现串联
        for (int j=0; j<m; j++) {
            int a=edge[j].a,b=edge[j].b,w=edge[j].w;
            dist[b] = min(dist[b],backup[a]+w);//只用上一次的
        }
    }
    if (dist[n] > 0x3f3f3f3f/2) cout << "impossible";
    else cout << dist[n];
}
int main() {
    cin >> n >> m >> k;
    for (int i=0; i<m; i++) {
        int a, b, w;
        cin >> a >> b >> w;
        edge[i] = {a, b, w};
    }
    bellman_ford();
    return 0;
}