HDU1548A strange lift广度优先搜索

http://acm.hdu.edu.cn/showproblem.php?pid=1548

A strange lift Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 34037    Accepted Submission(s): 12153   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 扫描二维码关注公众号，回复： 4448942 查看本文章 5 1 5 3 3 1 2 5 0   Sample Output 3

#include<bits/stdc++.h>
using namespace std;
int n,b,e;
int num[206];
int vis[206];
struct node
{
    int x;
    int step;//结构体储存x和step数
};
bool check(int x)
{
    if(x<=0||x>n)
    {
        return 1;
    }
    return 0;//定义check函数
}
int bfs()
{
    queue<node> q;
    node a,next;//定义两个node结构体
    int i;
    a.x=b;
    a.step=0;
    vis[b]=1;
    q.push(a);//放入一个结构体
    while(!q.empty())
    {
        a=q.front();
        q.pop();//取出
        if(a.x==e)
        {
            return a.step;
        }//终点判断
        for(int i=-1; i<=1; i+=2)
        {
            next=a;
            next.x+=i*num[next.x];
            if(check(next.x)||vis[next.x])
                continue;
            vis[next.x]=1;
            next.step++;
            q.push(next);
        }//向上向下走
    }
    return -1;
}
int main()
{
    int i,j;
    while(~scanf("%d",&n),n)
    {
        scanf("%d%d",&b,&e);
        for(i = 1; i<=n; i++)
            scanf("%d",&num[i]);
        memset(vis,0,sizeof(vis));
        printf("%d\n",bfs());
    }
    return 0;
}