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
題意：輸入n，a，b，和長度為n的k數組。共有n層電梯，求從a層到b層至少按幾次電梯。電梯只能按向上或者向下，在i層按上則上升ki層，按下則下降ki層，電梯不能到達1層以下，也不能到達n層以上。
思路：這道算是廣搜的基礎題了，深搜我試了試，超時。已經走過的樓層就不用走了，如果沒有可以走的樓層了還沒有到達目標點，那么就是沒有辦法到達。
代碼如下：

#include<bits/stdc++.h> #define ll long long #define Pll pair<int,int> using namespace std;const int maxx=2e2+10; int a[maxx],vis[maxx]; int n,s,e;inline int bfs() {vis[s]=1;queue<Pll> q;q.push(make_pair(s,0));while(q.size()){Pll u=q.front();q.pop();if(u.first==e) return u.second;int rr=u.first+a[u.first];if(rr<=n&&vis[rr]==0) vis[rr]=1,q.push(make_pair(rr,u.second+1));rr=u.first-a[u.first];if(rr>=1&&vis[rr]==0) vis[rr]=1,q.push(make_pair(rr,u.second+1));}return -1; } int main() {while(scanf("%d",&n),n){scanf("%d%d",&s,&e);memset(vis,0,sizeof(vis));for(int i=1;i<=n;i++) scanf("%d",&a[i]);printf("%d\n",bfs());}return 0; }

努力加油a啊，(^o)/~

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