hdu 1548(2013.9.15周赛F题BFS)

A strange lift

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID:1548
64-bit integer IO format: %I64d      Java class name: Main

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 53 3 1 2 50

Sample Output

3

题意：题意应该看得懂，不用解释了。刚开始不知道用的是BFS，让我用DFS弄了好久都RE，后面队友提醒了我，唉，实在是汗颜啊，以前挺怕STL的，以前只是大概看了一下，没有全弄明白，导致周赛都不知道怎么用，一点意识都没有，正好趁选修课老师去出差了偷空看了STL，才初步搞明白STL中的用法，而且还只是初步的，希望在一周或者一个月之后能运用自如，不然这些不懂就完蛋了，和别人的差距就更加大了，努力吧，这是简单的BFS，还行，终于是弄懂了一点皮毛……

#include<iostream>#include<cstring>#include<queue>using namespace std;int n,a,b,c[205],d[205];int bfs(){    int t,flag=0;    queue<int>q;    q.push(a);    d[a]=1;    while(!q.empty())    {        t=q.front();        q.pop();        if(t==b){flag=1;break;}        if(t-c[t]>0&&!d[t-c[t]])        {            q.push(t-c[t]);            d[t-c[t]]=d[t]+1;        }        if(t+c[t]<=n&&!d[t+c[t]])        {            q.push(t+c[t]);            d[t+c[t]]=d[t]+1;        }    }    if(flag) return d[b]-1;        else return -1;}int main(){    while(cin>>n&&n)    {        cin>>a>>b;        for(int i=1;i<=n;i++)            cin>>c[i];        memset(d,0,sizeof(d));        cout<<bfs()<<endl;}return 0;}