广度优先搜索：A strange lift HDU - 1548

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本文介绍了一个基于广度优先搜索（BFS）的算法，用于解决一个特殊电梯问题：给定每层楼电梯可上下的层数，计算从任意起点到终点所需的最少按钮按压次数。该算法通过遍历所有可能的路径来寻找最优解。

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
问题描述
第一行输入N：表示楼层数，A，B表示要从A走到B。
接下来的N个数表示每一层可以上下移动动的层数。
求最少需要移动多少次能从A走到B，如果无法走到则输出-1
问题分析
从A开始进行广度优先搜索，将能够到达的楼层进行标记，并加入到队列中，并且将步数加一，。
取队头进行和A相同的操作，直到队列为空，或者找到B。注意要将队头弹出并且不搜索已经标记过的楼层。
代码实现

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
class nod{
public :
    int ki;
    int step;//到达这一层已经走过的步数。
    nod():ki(0),step(0){}
}result[201];
int k_temp[201];
int bfs(const int B,const int N,queue<int> &que){
    while(!que.empty()){
        int temp=que.front();
        que.pop();
        if(temp==B)return result[temp].step;
        int count_A=temp+result[temp].ki;
        if(k_temp[count_A]==0&&count_A<N+1){
            que.push(count_A);
            k_temp[count_A]=1;
            result[count_A].step=result[temp].step+1;
        }
        int count_B=temp-result[temp].ki;
        if(k_temp[count_B]==0&&count_B>0){
            que.push(count_B);
            k_temp[count_B]=1;
            result[count_B].step=result[temp].step+1;
        }
    }
    return -1;
}
int main(){
    int N,A,B;
    while(scanf("%d",&N)){
        if(N==0)break;
        scanf("%d %d",&A,&B);
        for(int i=1;i<N+1;i++){
            scanf("%d",&result[i].ki);
            result[i].step=0;
            k_temp[i]=0;
        }
        queue<int> que;
        que.push(A);
        k_temp[A]=1;
        printf("%d\n",bfs(B,N,que));
    }
    return 0;
}