A strange lift
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 122 Accepted Submission(s) : 22
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.<br>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"?<br>
Input
The input consists of several test cases.,Each test
case contains two lines.<br>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.<br>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
3
简单题意：
给出N个楼梯up or down的参数， 和所在的楼层，和目标楼层，比如 在1楼按up那么会上升1楼的参数层，求到达目的楼层的最小次数
思路分析：
有目标层数，明显BFS。
# include <iostream>
# include <fstream>
# include <cstring>
# include <queue>
using namespace std;
int n, a, b;
int floor[1000];
int is[1000];
struct Info
{
int k;
int jishu;
};
int bfs();
int main()
{
//fstream cin("aaa.txt");
while(cin >> n)
{
if(n == 0)
break;
cin >> a >> b;
memset(is, 0, sizeof(is));
for(int i = 1; i <= n; i++)
cin >> floor[i];
int jishu = bfs();
cout << jishu << endl;
}
return 0;
}
int bfs()
{
is[a] = 1;
queue <Info> Q;
Info info;
info.k = a;
info.jishu = 0;
Q.push(info);
Info t, tep;
while(!Q.empty())
{
t = Q.front();
Q.pop();
if(t.k == b)
{
return t.jishu;
}
for(int i = -1; i <= 1; i++)
{
tep.k = t.k + i * floor[t.k];
if(tep.k >= 1 && tep.k <= n && !is[tep.k])
{
is[tep.k] = 1;
tep.jishu = t.jishu + 1;
Q.push(tep);
}
}
}
return -1;
}