专题测试 搜索 A - The Way to Home

1. 题目
   

   A frog lives on the axis Ox and needs to reach home which is in the point n. She starts from the point 1. The frog can jump to the right at a distance not more than d. So, after she jumped from the point x she can reach the point x + a, where a is an integer from 1 to d.

   For each point from 1 to n is known if there is a lily flower in it. The frog can jump only in points with a lilies. Guaranteed that there are lilies in the points 1 and n.

   Determine the minimal number of jumps that the frog needs to reach home which is in the point n from the point 1. Consider that initially the frog is in the point 1. If the frog can not reach home, print -1.

   Input

   The first line contains two integers n and d (2 ≤ n ≤ 100, 1 ≤ d ≤ n - 1) — the point, which the frog wants to reach, and the maximal length of the frog jump.

   The second line contains a string s of length n, consisting of zeros and ones. If a character of the string s equals to zero, then in the corresponding point there is no lily flower. In the other case, in the corresponding point there is a lily flower. Guaranteed that the first and the last characters of the string s equal to one.

   Output

   If the frog can not reach the home, print -1.

   In the other case, print the minimal number of jumps that the frog needs to reach the home which is in the point n from the point 1.

   Examples
   Input

   8 4
   10010101

   Output

   2

   Input

   4 2
   1001

   Output

   -1

   Input

   8 4
   11100101

   Output

   3

   Input

   12 3
   101111100101

   Output

   4

   Note

   In the first example the from can reach home in two jumps: the first jump from the point 1 to the point 4 (the length of the jump is three), and the second jump from the point 4 to the point 8 (the length of the jump is four).

   In the second example the frog can not reach home, because to make it she need to jump on a distance three, but the maximum length of her jump equals to two.

2. 思路
   签到题，本着尊重标题“专题搜索”的态度我写了个广搜，实际上可以用贪心的思想做。因为按题意青蛙能向右跳也只能向右跳，所以每次青蛙尽可能地跳最远就可以了，能到家那答案一定是最优的，没法跳回家则输出-1
3. 代码
   

   #include<cstdio>
   #include<cstring>
   #include<cmath>
   #include<algorithm>
   #include<queue>
   using namespace std;
   
   int t,n,d,x;
   int vis[333];
   int step[333];
   int map[333];
   int main()
   {
       scanf("%d%d",&n,&d);
       for(int i=1;i<=n;i++)
       {
           scanf("%1d",&map[i]);
       }
       queue<int>qx;
       qx.push(1);
       vis[1]=1;
       step[1]=0;
       step[n]=-1;
       while(!qx.empty())
       {
           x=qx.front(),qx.pop();
           if(x==n)
           {
               break;
           }
           for(int i=1;i<=d;i++)
           {
               if(map[x+i]==1&&vis[x+i]==0)
               {
                   qx.push(x+i);
                   step[x+i]=step[x]+1;
                   vis[x+i]=1;
               }
           }
       }
       //memset(vis,0,sizeof(vis));
       printf("%d\n",step[n]);
       return 0;
   }