hdu 1548 A strange lift 宽搜bfs+优先队列

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1548

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"?

题意描述：n层楼（1<=n<=200），每层楼一个数字Ki和两个按钮UP和DOWN，如果你在A层楼，A层楼的数字Ka=3，你可以按下UP按钮到A+Ka层楼，按下DOWN按钮到A-Ka层楼（前提是A+Ka和A-Ka都在n的范围里）。现在你在A层楼，要到目的地B层楼，问按下按钮的最少次数。

算法分析：直接bfs暴搜即可，由于涉及到最少次数问题，我习惯性的用到了优先队列处理。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<cmath>
 6 #include<algorithm>
 7 #include<queue>
 8 #define inf 0x7fffffff
 9 using namespace std;
10 const int maxn=200+10;
11 const int M = 40000+10;
12 
13 int n,A,B;
14 int an[maxn];
15 struct node
16 {
17     int cnt,id;
18     int value;
19     friend bool operator < (node a,node b)
20     {
21         return a.cnt > b.cnt;
22     }
23 }cur,tail;
24 
25 int vis[maxn];
26 int bfs()
27 {
28     priority_queue<node> Q;
29     cur.id=A ;cur.cnt=0 ;
30     cur.value=an[A];
31     Q.push(cur);
32     memset(vis,0,sizeof(vis));
33     vis[A]=1;
34     int count=0;
35     while (!Q.empty())
36     {
37         cur=Q.top() ;Q.pop() ;
38         if (cur.id==B) return cur.cnt;
39 
40         //cout<<cur.id<<" "<<an[cur.id]<<" "<<cur.cnt<<endl;
41         tail.id=cur.id+an[cur.id];
42         if (tail.id>=1 && tail.id<=n && !vis[tail.id])
43         {
44             tail.value=an[tail.id];
45             tail.cnt=cur.cnt+1;
46             vis[tail.id]=1;
47             Q.push(tail);
48         }
49 
50         tail.id=cur.id-an[cur.id];
51         if (tail.id>=1 && tail.id<=n && !vis[tail.id])
52         {
53             tail.value=an[tail.id];
54             tail.cnt=cur.cnt+1;
55             vis[tail.id]=1;
56             Q.push(tail);
57         }
58     }
59     return -1;
60 }
61 
62 int main()
63 {
64     while (scanf("%d",&n)!=EOF && n)
65     {
66         scanf("%d%d",&A,&B);
67         for (int i=1 ;i<=n ;i++) scanf("%d",&an[i]);
68         printf("%d
",bfs());
69     }
70     return 0;
71 }