HDU1548:A strange lift
A strange lift
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 64 Accepted Submission(s) : 29
Font: Times New Roman | Verdana | Georgia
Font Size: ← →
Problem Description
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 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
Sample Input
5 1 5
3 3 1 2 5
0
Sample Output
3
#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<climits>
using namespace std;
int a,b,ans,i,n;
int f[];
int step[];
void dfs(int floor,int k)
{
if (floor==b)
{
if (ans==-) ans=k;
else ans=min(ans,k);
return;
}
if (floor>n || floor<) return;
if (k>=step[floor]) return;
step[floor]=k;
dfs(floor+f[floor],k+);
dfs(floor-f[floor],k+);
return;
}
int main()
{
while(scanf("%d",&n),n)
{
scanf("%d%d",&a,&b);
for(i=;i<=n;i++)
{
scanf("%d",&f[i]);
step[i]=INT_MAX;
}
ans=-;
dfs(a,);
printf("%d\n",ans);
}
return ;
} /*bfs: 第二种方法
#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std; int a,b,ans,i,n;
int f[205];
int vis[205];
struct node
{
int floor,step;
};
int check(int a)
{
if(a<1 || a>n || vis[a]) return 0;
return 1;
}
int main()
{
while(scanf("%d",&n),n)
{
scanf("%d%d",&a,&b);
for(i=1;i<=n;i++)
scanf("%d",&f[i]);
if (a==b) {printf("0\n"); continue;}
memset(vis,0,sizeof(vis));
queue<node>s;
node p;
p.floor=a;
p.step=0;
s.push(p);
vis[a]=1;
ans=-1;
while(!s.empty())
{
p=s.front();
s.pop();
int x=p.floor+f[p.floor];
if (check(x))
{
vis[x]=1;
node q;
q.floor=x;
q.step=p.step+1;
s.push(q);
if(x==b) {ans=q.step;break; }
}
x=p.floor-f[p.floor];
if (check(x))
{
vis[x]=1;
node q;
q.floor=x;
q.step=p.step+1;
s.push(q);
if(x==b) {ans=q.step;break; }
}
}
printf("%d\n",ans);
}
return 0;
}*/
/*
之前用递归和bfs做的,这次用的是dijkstra
*/
#include <iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
const int inf=0x7fffffff;
int n,m,st,ed;
int vis[],dis[],mp[][];
int dijkstra()
{
memset(vis,,sizeof(vis));
for(int i=;i<=n;i++) dis[i]=mp[st][i]==?inf:;
vis[st]=;
for(int i=;i<n;i++)
{
int mindis=inf,k;
for(int j=;j<=n;j++)
if (!vis[j] && mindis>dis[j]) mindis=dis[j],k=j;
vis[k]=;
if (k==ed) return dis[k];
for(int j=;j<=n;j++)
if (!vis[j] && mp[k][j])
dis[j]=min(dis[j],dis[k]+);
}
return -;
}
int main()
{
while(scanf("%d",&n) && n!=)
{
scanf("%d%d",&st,&ed);
memset(mp,,sizeof(mp));
for(int i=;i<=n;i++)
{
int x;
scanf("%d",&x);
if(i+x<=n) mp[i][i+x]=;
if(i-x>) mp[i][i-x]=;
}
if (st==ed) {printf("0\n");continue;}
printf("%d\n",dijkstra());
}
return ;
}
HDU1548:A strange lift的更多相关文章
- HDU1548——A strange lift(最短路径:dijkstra算法)
A strange lift DescriptionThere is a strange lift.The lift can stop can at every floor as you want, ...
- HDU-1548 A strange lift(单源最短路 或 BFS)
Problem Description There is a strange lift.The lift can stop can at every floor as you want, and th ...
- Hdu1548 A strange lift 2017-01-17 10:34 35人阅读 评论(0) 收藏
A strange lift Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other) Tota ...
- hdu1548 A strange lift(bfs 或Dijkstra最短路径)
#include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #d ...
- HDU 1548 A strange lift (最短路/Dijkstra)
题目链接: 传送门 A strange lift Time Limit: 1000MS Memory Limit: 32768 K Description There is a strange ...
- HDU 1548 A strange lift (bfs / 最短路)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1548 A strange lift Time Limit: 2000/1000 MS (Java/Ot ...
- A strange lift
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission( ...
- bfs A strange lift
题目:http://acm.hdu.edu.cn/showproblem.php?pid=1548 There is a strange lift.The lift can stop can at e ...
- hdu 1548 A strange lift
题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1548 A strange lift Description There is a strange li ...
随机推荐
- LeetCode OJ 31. Next Permutation
Implement next permutation, which rearranges numbers into the lexicographically next greater permuta ...
- Excel教程(7) - 工程函数
BESSELI 用途:返回修正 Bessel 函数值,它与用纯虚数参数运算 时的 Bessel 函数值相等. 语法:BESSELI(x,n) 参数:X 为参数值.N 为函数的阶数.如果 n 非 ...
- 手机下的ev.pageX无效
把 ev.pageX 换成 e.originalEvent.targetTouches[0].pageX; 例子: var start_x, start_y, end_x, end_y, ...
- 解决MOSS列表匿名访问问题
//匿名 http://blog.csdn.net/yang_5/article/details/5214293 这是发布以后MOSS网站,如果你一开始在内部都没有启用匿名访问,你设置这个是不启作用 ...
- sharepoint2010配置个人网站的offical方法 来自Jetluning的专栏
Configuring My Site in SharePoint 201 SharePoint My Sites are commonly referred to as “Facebook fo ...
- Nginx正向代理让无法直接上网的机器通过代理上网
Nginx正向代理让无法直接上网的机器通过代理上网 在阿里云平台买了几台ECS.但是只要其中一台开通了公网.由于要初始化系统环境,需要网络安装相关依赖. Nginx正向代理配置: 一.Nginx 正向 ...
- 证明:一个整数a若不能被6整除,则a2+24必能被24整除。(整除理论,1.1.4)
证明:一个整数a若不能被6整除,则a2+24必能被24整除. 证明: 因为,a不能被6整除 所以,a不可以同时被2和3整除 所以,a一定是一个奇数, 所以,令a=2k+1,k是整数: 又因为,a2+2 ...
- 10. Regular Expression Matching
Implement regular expression matching with support for '.' and '*'. '.' Matches any single character ...
- Meta标签中的属性及含义
一.Meta标签中的format-detection属性及含义 format-detection翻译成中文的意思是“格式检测”,顾名思义,它是用来检测html里的一些格式的,那关于meta的forma ...
- iOS中的触摸事件,手势识别,摇晃事件等
在iOS中,事件可以划分为以下几类: 1.触摸事件:通过触摸,手势进行触发(手指点击.缩放等) 2.运动事件:通过加速器触发(例如手机晃动) 3.远程控制事件:通过其他远程设备触发(例如耳机控制按钮) ...