acm 2 1013 A strange lift
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1.1013
2.
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"?
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.
5 1 53 3 1 2 50
3
3.电梯每层有一个数字,每个数字不相同,若第n层有个数字k,那么这一层只能上k层或下k层,当然了不能低于1层或高于n层,给定起点与终点,求出最少要按几次键
4.这道题用搜索的思想可以做出,广度搜索
5.#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 t ;
}n1,n2,m;
int c[2000],ww[2000];
int flag ;
int n , a , b ;
int main( )
{
while( scanf("%d" , &n ) , n )
{
if( n== 0 ) break ;
cin>>a>>b;
for( int i= 1; i<= n ; i++ )
{ cin>>ww[i] ; c[i] = 0 ; }
flag = 0 ; //初始为0
n1.x = a ; n1.t = 0 ; //把a开始所在楼层赋给结构体, 所走步数为0
queue<node>Q;
Q.push(n1);
c[n1.x] = 1 ;
while( !Q.empty() )
{
m = Q.front() ;
Q.pop();
if( m.x == b )
{ flag = 1 ; break ; }
n1.x = m.x - ww[m.x];
n2.x = m.x + ww[m.x];
if( n1.x>0 && n1.x<= b && !c[n1.x] )
{
n1.t = m.t + 1 ;
c[n1.x] = 1 ;
Q.push(n1);
}
if( n2.x>0 && n2.x<= b && !c[n2.x] )
{
n2.t = m.t+1 ;
c[n2.x] = 1 ;
Q.push(n2);
}
}
if( flag ) cout<<m.t<<endl;
else cout<<"-1"<<endl;
}
return 0;
}
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