HDU 1548 A Strange lift DIJKSTRA+SPFA algorithm AC

A Strange LiftTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 13777 Accepted Submission (s): 5254


Problem Descriptionthere is a strange lift. The lift can stop can at every floor as you want, and there are a number ki (0 <= ki <= N) on every floor. The lift has just, buttons:up, and down. When you are at floor i,if Press the button "up", you'll go up Ki floor,i.e,you'll go to the i+ki th floor,as the SAM E, if you press the button ' down ', you'll go down Ki floor,i.e,you'll go to the i-ki th floor. Of course, the lift can ' t go up high than n,and can ' t go down lower than 1. For example, there are a buliding with 5 floors, and K1 = 3, K2 = 3,K3 = 1,k4 = 2, K5 = 5.Begining from the 1st Floor,you Can press the button "up", and you'll go up to the 4th Floor,and if your press the button "down", the lift can ' t does it, be Cause it can ' t go down to the-2nd floor,as you know, the-2nd floor isn ' t exist.
Here comes the problem:when is on floor a,and you want to go to floor b,how many times at least he have to press the Button ' up ' or ' down '?

Inputthe input consists of several test Cases.,each test case contains both lines.
The first line contains three integers n, a, B (1 <= n,a,b <=) which describe above,the second line consist N int Egers k1,k2,.... kn.
A single 0 Indicate the end of the input.
Outputfor each case of the input output a interger, the least times you had to press the button when you were on floor A,and y ou want to go to floor b.if your can ' t reach floor b,printf "-1".
Sample Input

5 1 53 3 1 2 50

Sample Output

3

Simple question. SPFA Code:

#include <cstdio> #include <deque> #define MAX #define INF 1000000000using namespace std; int Graph[max][ma X]; int T[max]; bool Visited[max]; int SPFA (int s, int d, int n) {int Dis[max], c[max];for (int i = 1; I <= n; ++i) { Dis[i] = INF; C[i] = 0; Visited[i] = false;}  Dis[s] = 0;d eque<int> que; Que.push_front (s); while (!que.empty ()) {int k = Que.front (); Que.pop_front (); Visited[k] = false; c[k]++; if (c[k]>n) {return-1;} for (int i = 1; I <= n; ++i) {if (Dis[i]>dis[k]+graph[k][i]) {dis[i] = Dis[k]+graph[k][i]; Visited[i] = true; if (!que. Empty ()) {if (Que.front () <=dis[i]) que.push_back (i); Elseque.push_front (i);} Else{que.push_front (i);}}} return dis[d];}  int main () {int n, a, B, while (~SCANF ("%d", &n) && N) {scanf ("%d%d", &a,&b); for (int i = 1; I <= N; ++i) {scanf ("%d", &t[i]);} for (int i = 0, I <= N; ++i) {for (int j = 0; J <= N; ++j) {graph[i][j] = INF;}} for (int i = 1; I <= n; ++i) {if (i+t[i]<=n) {Graph[i][i+t[i]] = 1;} if (i-t[i]>0) {Graph[i][i-t[i]] = 1;}} if (b<1|b>n) {puts ("-1"); continue;} int ans = SPFA (a,b,n), if (ans >= INF) puts ("1"); elseprintf ("%d\n", ans);} return 0;}

Dijkstra Code:

#include <stdio.h> #define MAX #define INF 1000000000int Graph[max][max]; int T[max], Dis[max];bool Visited[max] void Dijkstra (int s, int n) {for (int i = 1; I <= n; ++i) {dis[i] = Graph[s][i]; Visited[i] = false;} Dis[s] = 0;//forget to write this line of code. Let me wrong into sb. Visited[s] = true; for (int i = 1; i < n; ++i) {int index =-1, min = inf;for (int j = 1; j <= N; ++j) {if (!visited[j] && dis[j]<min) {index = j; min = Dis[j];}} if (index = =-1) {return;} Visited[index] = true; for (int j = 1; j <= N; ++j) {if (!visited[j] && dis[j]>dis[index]+graph[index][j]) {di S[J] = Dis[index]+graph[index][j];}}}  int main () {int n, a, B, while (~SCANF ("%d", &n) && N) {scanf ("%d%d", &a,&b); for (int i = 1; I <= N; ++i) {scanf ("%d", &t[i]);} for (int i = 0, I <= N; ++i) {for (int j = 0; J <= N; ++j) {graph[i][j] = INF;}} for (int i = 1; I <= n; ++i) {if (i+t[i]<=n) {Graph[i][i+t[i]] = 1;} if (i-t[i]>0) {Graph[i][i-t[i]] = 1;}} if (b<1|b>n) {pUTS ("-1"); continue;} Dijkstra (A,n), if (Dis[b] >= INF) puts ("1"); elseprintf ("%d\n", Dis[b]);} return 0;}

with June

HDU 1548 A Strange lift DIJKSTRA+SPFA algorithm AC