广度优先搜索——HDU1548

最新推荐文章于 2024-07-19 14:00:00 发布

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#搜索

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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：计算从楼层A到楼层B所需的最少按钮按压次数。电梯在每层楼有一个特定的跳跃数值，只能上跳或下跳该数值。通过使用队列和标记已访问楼层的方法，确保了找到最短路径。

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

5 1 5
3 3 1 2 5
0

Sample Output

简单BFS
注意是有向图

    #include <iostream>  
    #include <stdio.h>  
    #include <memory.h>  
    #include <queue>  
    using namespace std;  

    int N, A, B;  
    int a[205];  
    bool map[205], flag;  

    struct node  
    {  
        int x, step;  
    }n1, n2, m;  

    int main()  
    {  
        int i;  
        while(scanf("%d", &N), N)  
        {  
            if(N == 0) break;  
            scanf("%d %d", &A, &B);  
            for(i = 1; i <= N; i++)  
            {  
                scanf("%d", &a[i]);  
                map[i] = false;  
            }  
            flag = false;  
            n1.x = A; n1.step = 0;  
            queue<node> Q;  
            Q.push(n1);  
            map[n1.x] = true;  
            while(!Q.empty())  
            {  
                m = Q.front();  
                Q.pop();  
                if(m.x == B)    //到达目标  
                {  
                    flag = true;  
                    break;  
                }  
                n1.x = m.x - a[m.x];  
                n2.x = m.x + a[m.x];  
                if(n1.x>0 && n1.x<=B && !map[n1.x]) //下去的  
                {  
                    n1.step = m.step + 1;  
                    map[n1.x] = true;   //标记  
                    Q.push(n1);  
                }  
                if(n2.x>0 && n2.x<=B && !map[n2.x]) //上去的  
                {  
                    n2.step = m.step + 1;  
                    map[n2.x] = true;   //标记  
                    Q.push(n2);  
                }  
            }  
            if(flag) printf("%d\n", m.step);  
            else printf("-1\n");  
        }  

        return 0;  
    }