标签:
Catch That Cow4
//此题是在一条水平线上追赶羊,分为三种情况 x-1 /x+1 /2*x 注意边界即可
#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
int n,start,stop;
int p[210];
int vis[210];
struct Node
{
int floor;
int step;
};
int bfs(int flo)
{
queue <Node> q;
memset(vis,0,sizeof(vis));
Node a;
a.floor=flo,a.step=0;
q.push(a);
while(!q.empty())
{
Node b=q.front();
q.pop();
vis[b.floor]=1;
if(b.floor==stop) return b.step;
for(int i=0;i<2;i++) //0 up 1 down
{
Node c=b;
if(i==0)
c.floor=c.floor+p[c.floor];
else
c.floor=c.floor-p[c.floor];
if(!vis[c.floor]&&c.floor>=1&&c.floor<=n)
{
c.step++;
q.push(c);
vis[c.floor]=1;
}
}
}
return -1;
}
int main()
{
while(~scanf("%d",&n)&&n)
{
scanf("%d%d",&start,&stop);
for(int i=1;i<=n;i++)
scanf("%d",&p[i]);
printf("%d\n",bfs(start));
}
return 0;
}版权声明:本文为博主原创文章,未经博主允许不得转载。
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原文地址:http://blog.csdn.net/a73265/article/details/46860357