HDU1548 A strange life BFS

传送门

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

Sample Output

3

题目大意：一个奇怪的电梯，就是你要从A坐到B，每层电梯都有一个按钮,按钮上有数字，数字表示可以向上或者向下下降几层.k[i]表示在第i层的数字是k[i]. 问最少按几次按钮。

分析：这种题目吧，一看最少的话，简单题目就是BFS。 每一步有两中状态，分别进队，直到找到答案为止，注意在选择向上还是向下的时候 一定要判断向上的是否超过n，向下是否比1小。还有就是巧妙的利用vis[]数组，既然标记了,也起到了储存答案的作用

#include<iostream>
#include<queue>
#include<cstring>
#include<cstdio>
using namespace std;

const int maxn=;

int vis[maxn];
queue <int> q;
int a,b,n;
int k[];

void bfs(int x){
    q.push(x);
    while (!q.empty()) {
        int xx =q.front();
        q.pop();
        int down = xx-k[xx];
        int up = xx+k[xx];
        if(up<=n&&!vis[up]){
            vis[up]=vis[xx]+;
            q.push(up);
        }
        if(down>=&&!vis[down]){
            vis[down] = vis[xx]+;
            q.push(down);
        }
        if(up==b||down == b) break;
    }
}

int main()
{
    while(scanf("%d",&n),n){
        memset(vis,,sizeof vis);
        memset(k,,sizeof k);
        scanf("%d %d",&a,&b);
        while(!q.empty()){
            q.pop();
        }
        vis[a] = ;
        for(int i =  ;i <= n;i++){
            scanf("%d",&k[i]);
        }
        bfs(a);
        cout << vis[b]- << endl;
    }
    return ;
}