hrbust 哈理工OJ 1498Elevator Trouble【BFS过】

最新推荐文章于 2019-06-11 22:34:12 发布

mengxiang000000

最新推荐文章于 2019-06-11 22:34:12 发布

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分类专栏：搜索文章标签： hrbust 1498 哈理工OJ 1498

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本文介绍了一道关于电梯路径规划的问题，通过使用广度优先搜索算法解决如何从当前楼层到达目标楼层的问题。给出了完整的AC代码实现。

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Elevator Trouble

Time Limit: 1000 MS

Memory Limit: 65535 K

Total Submit: 108(34 users)

Total Accepted: 43(33 users)

Rating:

Special Judge: No

Description

    You are on your way to your first job interview as a program tester, and you are already late. The interview is in a skyscraper and you are currently in floor s, where you see an elevator. Upon entering the elvator, you learn that it has only two buttons,
 marked “ UP u" and DOWN d". You conclude that the UP-button takes the elevator u floors up(if there aren't enough floors, pressing the UP-botton does nothing, or at least so you assume), whereas the DOWN-button takes you d stories down (or none if there aren't
 enough). Knowing that the interview is at floor g, and that there are only f floors in the building, you quickly decide to write a program that gives you the amount of button pushes you need to perform. If you simply cannot reach the correct floor, your program
 halts with the message "use the stairs".

     Given input f, s, g, u and d(floors, start, goal, up, down), find the shortest sequence of button presses you must press in order to get fromstog, given a building of f floors, or output "use the stairs" if you cannot get from s to g by the given elevator.

Input

There are several test cases. For each test case, the input will consist of one line, namely f s g u d, where 1 <= s,g<=f<=1000000 and 0<=u,d<=1000000. The floors are one-indexed, i.e. if there are 10 stories,s and g be in [1,10].

Output

For each test case,You must reply with the minimum numbers of pushes you must make in order to get from

s to g, or output use the stairsif it is impossible given the con guration of the elvator.

Sample Input

1 10 2 1

2 1 1 0

Sample Output

6

use the stairs

Source

NCPC2011

一道并不是很难的搜索题，不需要剪枝、暴力搜一发就行了、

这个题和hdu 1548是一模一样的题目、大家有兴趣可以做做这个题目、

思路：从起点出发，两种走法，一种向上，一种向下，走到终点输出当前方法的步数。如果一直没有输出，就输出use the stairs

这里直接上AC代码：

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
struct floor
{
    int pos,output;
}now,nex;
int f,s,g,u,d;
int vis[1000005];
void bfs(int x)
{
    memset(vis,0,sizeof(vis));
    vis[x]=1;
    now.pos=x;
    now.output=0;
    queue<floor>s;
    s.push(now);
    while(!s.empty())
    {
        now=s.front();
        if(now.pos==g)
        {
            printf("%d\n",now.output);
            return ;
        }
        s.pop();
        for(int i=0;i<2;i++)
        {
            if(i==0)
            {
                nex.pos=now.pos+u;
                nex.output=now.output+1;
                if(nex.pos>=1&&nex.pos<=f&&vis[nex.pos]==0)
                {
                    vis[nex.pos]=1;
                    s.push(nex);
                }
            }
            if(i==1)
            {
                nex.pos=now.pos-d;
                nex.output=now.output+1;
                if(nex.pos>=1&&nex.pos<=f&&vis[nex.pos]==0)
                {
                    vis[nex.pos]=1;
                    s.push(nex);
                }
            }
        }
    }
    printf("use the stairs\n");
    return ;
}
int main()
{
    while(~scanf("%d%d%d%d%d",&f,&s,&g,&u,&d))
    {
        bfs(s);
    }
}