AcWing 853. 有边数限制的最短路
原题链接
简单
作者:
no_one
,
2022-08-06 16:06:30
,
所有人可见
,
阅读 120
复习bellman_ford
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int N = 520;
int n, m, k;
int d[N];
int back_up[N];
struct a
{
int a, b, c;
}edges[10010];
int bellman_ford()
{
memset(d, 0x3f, sizeof d);
d[1] = 0;
for(int i = 0; i < k; i ++)
{
memcpy(back_up, d, sizeof d);
for(int j = 0; j < m; j ++)
{
auto e = edges[j];
d[e.b] = min(d[e.b], back_up[e.a] + e.c);
}
}
if(d[n] > 0x3f3f3f3f / 2) return -0x3f3f;
return d[n];
}
int main()
{
cin >> n >> m >> k;
for(int i = 0; i < m; i ++)
{
int a, b, c;
cin >> a >> b >> c;
edges[i] = {a, b, c};
}
int t = bellman_ford();
if(t == -0x3f3f) puts("impossible");
else cout << t << endl;
return 0;
}