HDU 1548 A strange lift(最短路&&bfs）

Posted 2020-09-30 十年换你一句好久不见

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

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 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"?

 

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

 

Sample Output

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;
}