HDOJ_ACM_A strange lift

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 is on floor A,and you want to go to floor B,how many times at least he havt 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

3

 

 Code

/*
I use Breadth-first algorithm, that is BFS, to solve this question.
when u use BFS, u know u should mark the count of traversal, this is the reason I use Struct.
when u get the begin and end of floor, firstly u should push the begin into the queue.
then u should judge if the up and down of the floor which is in the top of the queue is available.
then u should pop from the queue, and push the available one into queue and mark the count until the queue is empty.
Note the count is difficult to mark so u should use Struct.
*/
#include <stdio.h>
#include <queue>      //in order to use queue
using namespace std;
#define N 200
int flag[N + 5];      //flag[i] = 1 means the floor has been arrived
int k[N + 5];         //k[i] is the number u can up or down
struct StepInfo       //in order to remember the step
{
    int floorNum;
    int step;
};
void main()
{
    int n, a, b, i, up, down, preStep;
    StepInfo s1, s2, s3;
    while (scanf("%d", &n) && n)
    {
        queue<StepInfo> q;   //note: u should declare the queue as local variable
        scanf("%d %d", &a, &b);  
        for (i = 1; i <= n; i++)
        {
            scanf("%d", &k[i]);
            flag[i] = 0;    // initialzing
        }
        //initialize the first stepInfo then push it into the queue
        s1.floorNum = a;
        s1.step = 0;
        q.push(s1);
        while (! q.empty())
        {
            preStep = q.front().step;
            //u find it
            if (q.front().floorNum == b)
                break;
            //mark that this floor was arrived
            flag[q.front().floorNum] = 1;
            //if u want to go up, firstly u should judge if the num is out of index
            up = q.front().floorNum + k[q.front().floorNum];
            //the same to down
            down = q.front().floorNum - k[q.front().floorNum];
            q.pop();
            //if the num is available, u can push it into the queue
            if (up >= 1 && up <= n && flag[up] != 1)
            {
                s2.floorNum = up;
                s2.step = preStep + 1;
                q.push(s2);
            }
            if (down >= 1 && down <= n && flag[down] != 1)
            {
                s3.floorNum = down;
                s3.step = preStep + 1;
                q.push(s3);
            }
        }
        //if u find it, it's sure that the queue is not empty
        if (!q.empty())
            printf("%d\n", preStep);
        else
            printf("-1\n");
    }
}

 Code

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