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=200+10;
const int M = 40000+10;

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=0 ;
    cur.value=an[A];
    Q.push(cur);
    memset(vis,0,sizeof(vis));
    vis[A]=1;
    int count=0;
    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>=1 && tail.id<=n && !vis[tail.id])
        {
            tail.value=an[tail.id];
            tail.cnt=cur.cnt+1;
            vis[tail.id]=1;
            Q.push(tail);
        }

        tail.id=cur.id-an[cur.id];
        if (tail.id>=1 && tail.id<=n && !vis[tail.id])
        {
            tail.value=an[tail.id];
            tail.cnt=cur.cnt+1;
            vis[tail.id]=1;
            Q.push(tail);
        }
    }
    return -1;
}

int main()
{
    while (scanf("%d",&n)!=EOF && n)
    {
        scanf("%d%d",&A,&B);
        for (int i=1 ;i<=n ;i++) scanf("%d",&an[i]);
        printf("%d
",bfs());
    }
    return 0;
}