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。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 #include<algorithm>
 #include<queue>
 #define inf 0x7fffffff
 using namespace std;
 const int maxn=+;
 const int M = +;

 int n,A,B;
 int an[maxn];
 struct node
 {
     int cnt,id;
     int value;
     friend bool operator < (node a,node b)
     {
         return a.cnt > b.cnt;
     }
 }cur,tail;

 int vis[maxn];
 int bfs()
 {
     priority_queue<node> Q;
     cur.id=A ;cur.cnt= ;
     cur.value=an[A];
     Q.push(cur);
     memset(vis,,sizeof(vis));
     vis[A]=;
     int count=;
     while (!Q.empty())
     {
         cur=Q.top() ;Q.pop() ;
         if (cur.id==B) return cur.cnt;

         //cout<<cur.id<<" "<<an[cur.id]<<" "<<cur.cnt<<endl;
         tail.id=cur.id+an[cur.id];
         if (tail.id>= && tail.id<=n && !vis[tail.id])
         {
             tail.value=an[tail.id];
             tail.cnt=cur.cnt+;
             vis[tail.id]=;
             Q.push(tail);
         }

         tail.id=cur.id-an[cur.id];
         if (tail.id>= && tail.id<=n && !vis[tail.id])
         {
             tail.value=an[tail.id];
             tail.cnt=cur.cnt+;
             vis[tail.id]=;
             Q.push(tail);
         }
     }
     return -;
 }

 int main()
 {
     while (scanf("%d",&n)!=EOF && n)
     {
         scanf("%d%d",&A,&B);
         for (int i= ;i<=n ;i++) scanf("%d",&an[i]);
         printf("%d\n",bfs());
     }
     return ;
 }