hdu 1548:A strange lift
2014-03-07 01:42
281 查看
一部电梯,能上能下。在第i层有一个k,电梯在i层只能上到第i + k或下到第i – k层,且不会超过第N层或低于第1层。问从A层到B层最少上升或下降几次。若无法到达则输出-1。
Spfa简单搞定。不过这题既视感好强。。。
Spfa简单搞定。不过这题既视感好强。。。
#include <cstdio> #include <cstring> #include <queue> #include <iostream> using namespace std; const int inf = 1 << 20 ; const int MAX = 205 ; int N , A , B ; int kk[MAX] ; int dis[MAX] ; bool had[MAX] ; queue <int> q ; void spfa() { for (int i = 1 ; i <= N ; i ++) { dis[i] = inf ; } dis[A] = 0 ; q.push(A) ; had[A] = 1 ; while (!q.empty()) { int t = q.front() ; q.pop() ; had[t] = 0 ; int t_p = t + kk[t] ; int t_m = t - kk[t] ; if (t_p <= N && dis[t_p] > dis[t] + 1) { dis[t_p] = dis[t] + 1 ; if (!had[t_p]) { had[t_p] = 1 ; q.push(t_p) ; } } if (t_m >= 1 && dis[t_m] > dis[t] + 1) { dis[t_m] = dis[t] + 1 ; if (!had[t_m]) { had[t_m] = 1 ; q.push(t_m) ; } } } } int main() { while (cin >> N) { if (!N) break ; cin >> A >> B ; for (int i = 1 ; i <= N ; i ++) { cin >> kk[i] ; } spfa() ; if (dis[B] == inf) cout << -1 << endl ; else cout << dis[B] << endl ; } return 0 ; }
相关文章推荐
- HDU 1548 A strange lift bfs 宽搜解析
- HDU 1548 A strange lift
- HDU 1548 A strange lift (Dijkstra)
- hdu 1548 A strange lift
- HDU 1548 A strange lift
- HDU 1548 A strange lift(BFS)
- HDU 1548 A strange lift (Dijkstra算法)
- HDU 1548 A strange lift
- HDU 1548 A strange lift
- HDU-1548 A strange lift(最短路[Spfa || BFS])
- HDU 1548 A strange lift-bfs
- hdu 1548 A strange lift (bfs)
- hdu 1548 A strange lift(最短路)
- Hdu1548 A strange lift【简单bfs】
- hdu 1548 A strange lift BFS 解法
- HDU 1548 A strange lift
- HDU 1548 A strange lift
- hdu 1548 A strange lift(迪杰斯特拉,邻接表)
- hdu 1548 A strange lift
- Hdu 1548 A strange lift(BFS)