HDU 1548 A strange lift .

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分类专栏： BFS HDU OJ 文章标签： c++ BFS

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本文探讨了一个具有特殊楼层跳跃功能的电梯模型，并通过BFS（宽度优先搜索）算法结合剪枝策略来解决从任意起点到达终点所需的最少按钮按压次数问题。输入包括楼层总数、起始和目标楼层及各楼层的跳跃距离。

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A strange lift
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3233 Accepted Submission(s): 1157

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
看清楚模型后， BFS+剪枝的力量是强大的{>0<}

3614188 2011-03-13 19:29:35 Accepted 1548 0MS 268K 855 B C++ 10SGetEternal{（。）（。）}!


#include<queue>
#include<iostream>
using namespace std;
int main()
{
 int i,n,a,b,flo;
 bool vis[201];
 int ki[201],step[201];
 while (scanf("%d",&n),n)
 {
  scanf("%d%d",&a,&b);
  queue<int> Q;
  for (i=1;i<=n;i++)
  {
   scanf("%d",&ki[i]);
   vis[i]=1;
   step[i]=(1<<31)-1;
  }
  Q.push(a);
  step[a]=0;
  while (!Q.empty())
  {
   flo=Q.front();
   if (flo==b) break;
   Q.pop();
   vis[flo]=0;
   if (flo+ki[flo]<=n && flo+ki[flo]>=1)
    if (vis[flo+ki[flo]] && step[flo]+1<=step[flo+ki[flo]]) 
    {
     step[flo+ki[flo]]=step[flo]+1;
     Q.push(flo+ki[flo]);
    }
   if (flo-ki[flo]<=n && flo-ki[flo]>=1)
    if (vis[flo-ki[flo]] && step[flo]+1<=step[flo-ki[flo]]) 
    {
     step[flo-ki[flo]]=step[flo]+1;
     Q.push(flo-ki[flo]);
    }
  }
  printf(flo==b?"%d/n":"-1/n",step[b]);
 }
 return 0;
}

BFS就系最短路ge神器……………………