HDU 1548 A strange lift(最短路&&bfs）

A strange lift

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26943 Accepted Submission(s): 9699


[align=left]Problem Description[/align]
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 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"?

[align=left]Input[/align]
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.

[align=left]Output[/align]
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".

[align=left]Sample Input[/align]

5 1 5
3 3 1 2 5
0

[align=left]Sample Output[/align]

3
最短路是有向图，不是无向图，一个电梯向上不过n，向下不过1。

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 0x3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
int dis[250],g[250][250],vis[250];
int a[250],n,x,y,k;
void init()
{
for(int i=1;i<=n;i++)
{
for(int j=1;j<i;j++)
{
g[i][j]=g[j][i]=INF;
}
g[i][i]=0;
}
}
void dij(int x)
{
for(int i=1;i<=n;i++)
{
dis[i]=g[x][i];
vis[i]=0;
}
vis[x]=1;
int minn,v=x;
for(int i=1;i<n;i++)
{
minn=INF;
for(int j=1;j<=n;j++)
{
if(!vis[j] && minn>dis[j])
{
minn=dis[j];
v=j;
}
}
if(minn==INF) break;
vis[v]=1;
for(int j=1;j<=n;j++)
{
if(!vis[j]) dis[j]=min(dis[j],dis[v]+g[v][j]);
}
}
}
int main()
{
while(scanf("%d",&n)&&n)
{
init();
scanf("%d%d",&x,&y);
for(int i=1;i<=n;i++)
{
scanf("%d",&k);
if(i-k>=1) g[i][i-k]=1;
if(i+k<=n) g[i][i+k]=1;
}
dij(x);
printf("%d\n",dis[y]==INF?-1:dis[y]);
}
return 0;
}

bfs

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 0x3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
int vis[250],x,n;
int dis[250],y,xx;
struct node
{
int x;
int step;
}ans,pos;
int bfs(int A,int B)
{
queue<node>q;
memset(vis,0,sizeof(vis));
vis[A]=1;
pos.x=A;pos.step=0;
q.push(pos);
while(!q.empty())
{
pos=q.front();
q.pop();
if(pos.x==B) return pos.step;
xx=pos.x+dis[pos.x];
if(x<=n && !vis[xx])
{
ans.x=xx;
ans.step=pos.step+1;
vis[xx]=1;
q.push(ans);
}
xx=pos.x-dis[pos.x];
if(x>=1 && !vis[xx])
{
ans.x=xx;
ans.step=pos.step+1;
vis[xx]=1;
q.push(ans);
}
}
return -1;
}
int main()
{
while(scanf("%d",&n)&&n)
{
scanf("%d%d",&x,&y);
for(int i=1;i<=n;i++)
{
scanf("%d",&dis[i]);
}
printf("%d\n",bfs(x,y));
}
return 0;
}