bellman-ford AcWing 853. Shortest path with side limit

bellman-ford AcWing 853. The shortest path with limited number of edges

Original link

AcWing 853. The shortest path with limited number of edges

Algorithm tags

shortest path Bellman-Ford

Ideas

From the meaning of the title There are negative edges , So we can't use Dijkstra
Due to the limitation of the number of sides
Therefore adopt Bellman-Ford
Bellman-Ford Algorithm

For paths with negative weight loops
There may be no shortest circuit
Such as below
1 to 5 shortest path non-existent Because every time in 2, 3, 4 Loop once , How many lengths of the road minus 1

Code

#include<bits/stdc++.h>
#define int long long
#define rep(i, a, b) for(int i=a;i<b;++i)
#define Rep(i, a, b) for(int i=a;i>b;--i)
using namespace std;
const int N = 10005;
// p Storage node ancestor node  d Store the distance between the current node and the ancestor node 
int lst[N], dis[N];
int n,m,k;
inline int read(){
   int s=0,w=1;
   char ch=getchar();
   while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
   while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
   return s*w;
}
void put(int x) {
    if(x<0) putchar('-'),x=-x;
    if(x>=10) put(x/10);
    putchar(x%10^48);
}
struct Edge{
    int a, b, c;
}edge[N];
int be(){
    memset(dis, 0x3f3f, sizeof dis);
    dis[1]=0;
    rep(i, 0, k){
        memcpy(lst, dis, sizeof dis);
        rep(j, 0, m){
            Edge e=edge[j];
            dis[e.b]=min(dis[e.b], lst[e.a]+e.c);
        }
    }
    if(dis[n]>0x3f3f3f3f/2){
        return -1;
    }else{
        return dis[n];
    }
}
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    n=read(), m=read(), k=read();
    rep(i, 0, m){
        int x=read(), y=read(), z=read();
        edge[i]={x, y, z};
    }
    if(be()==-1){
        puts("impossible");
    }else{
        printf("%lld", dis[n]);
    }
    return 0;
}

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