HDU1548A strange lift（bfs）

最新推荐文章于 2021-12-18 21:01:38 发布

ling_wang

最新推荐文章于 2021-12-18 21:01:38 发布

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本文探讨了一种特殊电梯模型的问题解决方法，该电梯在每一层可以向上或向下移动特定的层数。通过使用广度优先搜索算法，我们找到了从任意起点到达终点所需的最少按钮按压次数。

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A strange lift
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

题意：
一个人要从某一层电梯到另一层电梯，每层电梯都只能上或者下特定层数，问你从某一层到另一层最少需要按多少次电梯按钮

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<string>
#include<cmath>
#include<queue>
#include<vector>
using namespace std;
int N, A, B;
const int maxn = 200 + 10;
bool vis[maxn];
int k[maxn];
int check(int x)
{
    if(x<=0 || x>N)
        return 1;
    return 0;
}
struct lift
{
    int f, step;
};
int bfs()
{
    int i;
    queue <lift> q;
    lift s, t;
    s.f = A;
    s.step = 0;
    vis[s.f] = 1;
    q.push(s);
    while(!q.empty())
    {
        s = q.front();
        q.pop();
        if(s.f == B){
            return s.step;
        }
        for(i=-1; i<=1; i+=2){
            t = s;
            t.f += i*k[t.f];
            if(check(t.f) || vis[t.f]){
                continue;
            }
            vis[t.f] = 1;
            t.step++;
            q.push(t);
        }
    }
    while(!q.empty())   q.pop();
    return -1;
}
int main()
{
    while(scanf("%d", &N), N)
    {
        scanf("%d %d", &A, &B);
        memset(vis, 0, sizeof(vis));
        for(int i=1; i<=N; i++){
            cin >> k[i];
        }
        int myans = bfs();
        printf("%d\n", myans);
    }
    return 0;
}