（HDU - 1548）A strange lift

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A strange lift

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Problem Description

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”?

Input

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.

Output

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”.

Sample Input

5 1 5
3 3 1 2 5
0

Sample Output

题目大意：一个电梯，每层楼都有一个按钮选择上楼还是下楼，当电梯在i层楼可以向上或向下走k[i]层到达i-k[i]或i+k[i]层楼，当然电梯所在楼层不能低于1也不能高于n。

思路：最短路或者bfs即可求解，因为我刚学bfs，所以用bfs进行求解，题目不难，属于基础的bfs，注意输入输出格式即可。

#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;

const int maxn=205;
bool vis[maxn];
int k[maxn][2];
int n,a,b;

struct node
{
    int x,step;
};

bool check(int x)
{
    return (x>=1&&x<=n);
}

int bfs()
{
    queue<node> q;
    node start,tmp,next;
    start.x=a;
    start.step=0;
    vis[a]=true;
    q.push(start);
    while(!q.empty())
    {
        tmp=q.front();
        q.pop();
        if(tmp.x==b) return tmp.step;
        for(int i=0;i<2;i++)
        {
            next.x=tmp.x+k[tmp.x][i];
            if(next.x==b) return tmp.step+1;
            if(check(next.x)&&!vis[next.x])
            {
                next.step=tmp.step+1;
                vis[next.x]=true;
                q.push(next);
            }
        }
    }
    return -1;//不能从a到达b返回-1 
}

int main()
{
    while(scanf("%d",&n)!=EOF&&n)
    {
        scanf("%d%d",&a,&b);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&k[i][0]);
            k[i][1]=-k[i][0];
        }
        memset(vis,false,sizeof(vis));
        printf("%d\n",bfs());
    }
    return 0;
}