2016SDAU編程練習二1013

原創

2020-02-22 08:02

A strange lift

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 is on floor A,and you want to go to floor B,how many times at least he havt 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

題意：有一個特別的電梯，第i層有一個對應的數字ki，對於第i層按上升鍵up可升上到i+k[i]層，按下降鍵down到達i-k[i]層，到達的樓層最高不能超過n層，最低不能小於1層。給你一個起點A和終點B，問最少要按幾次上升鍵或者下降鍵到達目的地。

思路：乍一看有點像貪心，BFS

感想：搜索好。。。。長。。。。。

AC代碼：

#include<iostream>
#include<string.h>
#include<set>
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<numeric>
#include<math.h>
#include<string.h>
#include<sstream>
#include<stdio.h>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<queue>
#include<iomanip>
#include<cstdio>
using namespace std;
struct node
{
int x ;
int y ;
}n1,n2,m;
int c[2000],d[2000]; //標記數組，和一個記錄樓層狀態的數組
int flag ; //記錄能否到達目標樓層
int n , a , b ; //n層，a,起始樓層，b，目的樓層
int main( )
{
while( scanf("%d" , &n ) , n )
{
if( n== 0 ) break ; //結束
cin>>a>>b;
for( int i= 1; i<= n ; i++ ) //初始化
{ cin>>d[i] ; c[i] = 0 ; }
flag = 0 ; //初始爲0
n1.x = a ; n1.y = 0 ; //把a開始所在樓層賦給結構體，所走步數爲0
queue<node>Q; //建立隊列Q
Q.push(n1); //入隊
c[n1.x] = 1 ; //標記已經走過
while( !Q.empty() ) //直到爲空爲止
{
m = Q.front() ; //把隊列中第一個值賦給結構體變量m
Q.pop(); //出隊
if( m.x == b ) //如果到達終點便退出
{ flag = 1 ; break ; }
n1.x = m.x - d[m.x]; //按向下的按鈕
n2.x = m.x + d[m.x]; //按向上的按鈕
if( n1.x>0 && n1.x<= n&& !c[n1.x] )//向下
{
n1.y = m.y + 1 ; //步數加一
c[n1.x] = 1 ; //標記
Q.push(n1);
}
if( n2.x>0 && n2.x<= n && !c[n2.x] )//向上
{
n2.y = m.y+1 ;
c[n2.x] = 1 ;
Q.push(n2);
}
}
if( flag ) cout<<m.y<<endl;
else cout<<"-1"<<endl;
}
return 0;
}