HDU-1548--A strange lift--(BFS,剪枝)
A strange lift
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
Total Submission(s): 945 Accepted Submission(s): 450 |
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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 is on floor A,and you want to go to floor B,how many times at least he havt 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".
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Sample Input
5 1 5 3 3 1 2 5 0 |
Sample Output
3 |
Recommend
8600
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做地图搜索做到恶心,终于来了道换种问法的,其实还是一样。。。。(汗。。。
这道题如果裸搜索,不剪枝的话,会超时
于是用一个vis数组标记一下此电梯是否访问到,如果访问到就无需加入到队列中(此时队列中的状态一定是从已访问的那一层得来的)
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<map> #include<iomanip> #include<queue> #define INF 0x7ffffff #define MAXN 220 using namespace std; const double eps=1e-10; int n,a,b; struct node { int f,step; }; int k[MAXN]; int vis[MAXN]; int bfs() { int x; queue<struct node> q; struct node pre,now; pre.f=a;pre.step=0; q.push(pre); vis[a]=1; while(!q.empty()){ pre=q.front(); q.pop(); if(pre.f==b) return pre.step; x=pre.f; now=pre; now.step++; if(x+k[x]<=n&&vis[x+x[k]]==0){ vis[x+x[k]]=1; now.f+=k[x]; q.push(now); now.f-=k[x]; } if(x-k[x]>=1&&vis[x-x[k]]==0){ vis[x-x[k]]=1; now.f-=k[x]; q.push(now); } } return -1; } int main() { #ifndef ONLINE_JUDGE freopen("data.in", "r", stdin); #endif std::ios::sync_with_stdio(false); std::cin.tie(0); int res; while(cin>>n&&n!=0){ memset(vis,0,sizeof(vis)); cin>>a>>b; for(int i=1;i<=n;i++){ cin>>k[i]; } res=bfs(); if(res==-1) cout<<-1<<endl; else cout<<res<<endl; } }