hdu 1548 A strange lift

[align=left]Problem Description[/align]
There is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 <= Ki <= N) on every floor.The lift have just two buttons: up and down.When you at floor i,if you press the button "UP" , you will
go up Ki floor,i.e,you will go to the i+Ki th floor,as the same, if you press the button "DOWN" , you will go down Ki floor,i.e,you will 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
is a buliding with 5 floors, and k1 = 3, k2 = 3,k3 = 1,k4 = 2, k5 = 5.Begining from the 1 st floor,you can press the button "UP", and you'll go up to the 4 th floor,and if you press the button "DOWN", the lift can't do it, because it can't go down to the -2
th floor,as you know ,the -2 th floor isn't exist.

Here comes the problem: when you are on floor A,and you want to go to floor B,how many times at least he has to press the button "UP" or "DOWN"?

 

[align=left]Input[/align]
The input consists of several test cases.,Each test case contains two lines.

The first line contains three integers N ,A,B( 1 <= N,A,B <= 200) which describe above,The second line consist N integers k1,k2,....kn.

A single 0 indicate the end of the input.
 

[align=left]Output[/align]
For each case of the input output a interger, the least times you have to press the button when you on floor A,and you want to go to floor B.If you can't reach floor B,printf "-1".
 

[align=left]Sample Input[/align]

5 1 5
3 3 1 2 5
0

 

[align=left]Sample Output[/align]
[align=left]3[/align]
[align=left]这道题初始想法是用BFS，后来发现原来dijkstra也可以做。[/align]
[align=left]具体做法就是将每一层电梯看作一个结点，如果层x有通向层y的路，则将x，y边的权值记为1，无路则为INF（构边时为单向边）然后dijkstra求从结点A到结点B的最短路径就[/align]
[align=left]ok了。[/align]

#include<iostream>
#include<cstring>
using namespace std;
#define INF 0x3f3f3f3f
int n,a,b;
int p[201];
int map[201][201];
int d[201];
int vis[201];
void dij(){
memset(map,0x3f,sizeof(map));
int i,j,v,min;
for(i=1;i<=n;i++){
map[i][i]=0;
vis[i]=0;
if(i+p[i]<=n)
map[i][i+p[i]]=1;
if(i-p[i]>=1)
map[i][i-p[i]]=1;

}
for(i=1;i<=n;i++){
d[i]=map[a][i];
}
for(i=1;i<=n;i++){
min=INF;
for(j=1;j<=n;j++){
if(vis[j]==0&&d[j]<min){
v=j;
min=d[j];
}

}vis[v]=1;
for(j=1;j<=n;j++){
if(vis[j]==0&&d[j]>d[v]+map[v][j])
d[j]=d[v]+map[v][j];
}

}if(d[b]<INF)
cout<<d[b]<<endl;
else
cout<<-1<<endl;

}
int main(){
int i,j;
while(cin>>n&&n!=0){
cin>>a>>b;
for(i=1;i<=n;i++){
cin>>p[i];
}
dij();

}

}

[align=left]
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