A strange lift_BFS
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.
3 3 1 2 5
0
【题意】给出n层,要从s到t,给出每层只能下或者上a[i],问至少经过多少次才能从s到t
【思路】经典BFS
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
const int N=;
int a[N];
int vis[N];
int n,s,t;
struct node
{
int x;
int step;
};
bool go(int x)
{
if(x<=||x>n) return false;
return true;
}
int bfs()
{
node now,next;
queue<node>qu;
now.x=s;
now.step=;
qu.push(now);
while(!qu.empty())
{
now=qu.front();
qu.pop();
if(now.x==t)
return now.step;
for(int i=-;i<=;i+=)
{
next=now;
next.x+=a[next.x]*i;
if(go(next.x)&&vis[next.x]==)
{
vis[next.x]=;
next.step++;
qu.push(next); }
}
}
return -;
}
int main()
{
while(~scanf("%d",&n),n)
{
scanf("%d%d",&s,&t);
for(int i=;i<=n;i++)
scanf("%d",&a[i]);
memset(vis,,sizeof(vis));
int ans=bfs();
printf("%d\n",ans);
}
return ;
}
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