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]

3
题目大意： 坐电梯是，输入在每一层都有一个数字Ki（0 <= Ki <= N）。电梯只有两个按钮：上和下。当你在楼层i，如果你按下“UP”按钮，你会上去Ki楼层，也就是说，你会去i + Ki楼层，如果你按下“DOWN”按钮，你会下去Ki楼，即，你会去i-Ki楼。当然，电梯不能上升高于N，并且不能下降到低于1.例如，具有5个楼层的楼层，并且k1 = 3，k2 = 3，k3 = 1，k4 = 2，k5 = 5.从第一层开始，你可以按下“UP”按钮，你会上到4楼，如果你按下“DOWN”按钮，电梯不能它，因为它不能下到-2楼，你知道，-2楼不存在。
这里的问题：当你在A楼，你想去B楼，至少要多少次按下“UP”或“DOWN”按钮？解题思路： 在这个题中，我们可以采用两种方法，一种是BFS，一种是最短路径算法（迪杰斯特拉等）。再用优先队列时，用bfs从起点开始存入队列开始搜，假设下一个点没有走过而且在1~n层之间，则压入队列。符合条件则输出步数。最短距离的如程序备注所示#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
//迪杰斯特拉法
const int inf = 1<<30; //最大值

int n;
int map[205][205]; //映射二维数组
int a[205],cnt; //
int vis[205],cast[205];

void Dijkstra(int s,int e)//迪杰斯特拉最短路径算法
{
int i,j,min,pos;
memset(vis,0,sizeof(vis)); //初始化vis数组
for(i = 0; i<n; i++) //起点位置向周边
{
cast[i] = map[s][i];
}

cast[s] = 0; //起点位置消耗0
vis[s] = 1; //起始位置访问点设置为1
for(i = 1; i<n; i++)
{
min = inf; //让最小值最大化
for(j = 0; j<n; j++)
{
if(cast[j]<min && !vis[j])//从位置开始向周边访问获取最短距离
{
pos = j;
min = cast[j];
}
}
if(min == inf) //如果这个最短距离是最大值，则退出
break;
vis[pos] = 1; //标志访问过的且是最短距离位置的那个点标志位1
for(j = 0; j<n; j++)
{
if(cast[pos]+map[pos][j]<cast[j] && !vis[j]) //
cast[j] = cast[pos]+map[pos][j];
}
}
}

int main()
{
int i,j,s,e,x,y;
while(~scanf("%d",&n),n)
{
scanf("%d%d",&s,&e);
s--,e--;
for(i = 0; i<n; i++)
for(j = 0; j<n; j++)
map[i][j] = inf;
for(i = 0; i<n; i++)
{
scanf("%d",&a[i]);
if(i+a[i]<n)
map[i][i+a[i]] = 1;
if(i-a[i]>=0)
map[i][i-a[i]] = 1;
}
Dijkstra(s,e);
printf("%d\n",cast[e]==inf?-1:cast[e]);
}

return 0;
}