HDOJ-1548 A strange lift -------BFS/dijkstra

A strange lift

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6572    Accepted Submission(s): 2443

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

 

  1 /* 功能Function Description:     HDOJ-1548    
  2    开发环境Environment:          DEV C++ 4.9.9.1
  3    技术特点Technique:
  4    版本Version:
  5    作者Author:                   可笑痴狂
  6    日期Date:                      20120821
  7    备注Notes:
  8    这题一看就是BFS题，但是老师让用最短路径做，谁知道测试数据太变态了用最短路径WA了半天
  9    用BFS时注意要访问标记一下，要不会MLE
 10 */
 11 /*
 12 //代码一：(用队列实现广搜)
 13 #include<cstdio>
 14 #include<queue>
 15 using namespace std;
 16 
 17 int button[205];
 18 struct node
 19 {
 20     int floor;
 21     int time;
 22 };
 23 
 24 int BFS(int a,int b,int n)
 25 {
 26     
 27     queue<node> q;
 28     bool visit[205];   //没有访问标记会MLE
 29     node tmp,s;
 30     memset(visit,false,sizeof(visit));
 31     visit[a]=true;
 32     s.floor=a;
 33     s.time=0;
 34     q.push(s);
 35     while(!q.empty())
 36     {
 37         s=q.front();
 38         q.pop();
 39         if(s.floor==b)
 40             return s.time;
 41         else
 42         {
 43             for(int i=1;i<=2;++i)
 44             {
 45                 if(i==1)
 46                     tmp.floor=s.floor+button[s.floor];  //向上
 47                 else                                    
 48                     tmp.floor=s.floor-button[s.floor];  //向下
 49                 if(tmp.floor<=0||tmp.floor>n||visit[tmp.floor])
 50                     continue;
 51                 else
 52                 {
 53                     tmp.time=s.time+1;
 54                     visit[tmp.floor]=true;
 55                     q.push(tmp);
 56                 }
 57             }
 58         }
 59     }
 60     return -1;
 61 }
 62 
 63 int main()
 64 {
 65     int n,a,b,i;
 66     while(scanf("%d",&n),n)
 67     {
 68         scanf("%d%d",&a,&b);
 69         for(i=1;i<=n;++i)
 70             scanf("%d",&button[i]);
 71         printf("%d\n",BFS(a,b,n));
 72     }
 73     return 0;
 74 }
 75 */
 76 
 77 //代码二：----dijkstra
 78 #include<stdio.h>
 79 #include<string.h>
 80 #define inf 0x3fffffff
 81 int map[205][205];
 82 
 83 int dijks(int s,int t,int n)
 84 {
 85     bool visit[205];
 86     int dis[205];
 87     int i,j,k,min;
 88     memset(visit,false,sizeof(visit));
 89     visit[s]=true;
 90     for(i=1;i<=n;++i)
 91         dis[i]=map[s][i];
 92     for(i=1;i<n;++i)
 93     {
 94         min=inf;
 95         k=1;
 96         for(j=1;j<=n;++j)
 97         {
 98             if(!visit[j]&&min>dis[j])
 99             {
100                 min=dis[j];
101                 k=j;
102             }
103         }
104         if(k==t||min==inf)
105             break;
106         visit[k]=true;
107         for(j=1;j<=n;++j)
108         {
109             if(!visit[j]&&dis[k]+map[k][j]<dis[j])
110                 dis[j]=dis[k]+map[k][j];
111         }
112     }
113     if(dis[t]==inf)
114         return -1;
115     else
116         return dis[t];
117 }
118 
119 int main()
120 {
121     int n,a,b,i,j;
122     int button[205];
123     while(scanf("%d",&n),n)
124     {
125         scanf("%d%d",&a,&b);
126         for(i=1;i<=n;++i)
127             scanf("%d",&button[i]);
128         if(a==b)    //注意这点，测试数据有点变态，超出楼层高度的数据
129         {
130             printf("0\n");
131             continue;
132         }
133         if(a>n||b>n)
134         {
135             printf("-1\n");
136             continue;
137         }
138         for(i=1;i<=n;++i)
139             for(j=1;j<i;++j)
140                 map[i][j]=map[j][i]=inf;
141         for(i=1;i<=n;++i)
142         {
143             if(i+button[i]<=n)
144                 map[i][i+button[i]]=1;
145             if(i-button[i]>0)
146                 map[i][i-button[i]]=1;
147         }
148         printf("%d\n",dijks(a,b,n));
149     }
150     return 0;
151 }