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 3 問題描述 第一行輸入N：表示樓層數，A，B表示要從A走到B。 接下來的N個數表示每一層可以上下移動動的層數。 求最少需要移動多少次能從A走到B，如果無法走到則輸出-1 問題分析 從A開始進行廣度優先搜索，將能夠到達的樓層進行標記，并加入到隊列中，并且將步數加一，。 取隊頭進行和A相同的操作，直到隊列為空，或者找到B。注意要將隊頭彈出并且不搜索已經標記過的樓層。 代碼實現