hdu 1548 A strange lift（迪杰斯特拉，邻接表）

A strange lift

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18723 Accepted Submission(s):
6926


[align=left]Problem Description[/align]
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"?

[align=left]Input[/align]
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.

[align=left]Output[/align]
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".

[align=left]Sample Input[/align]

5 1 5

3 3 1 2 5

0

[align=left]Sample Output[/align]

3

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最短路的模板题，代码还是很好理解的。

题意：一个特别的电梯，按up可升上k[i]层，到大i+k[i]层，down则到达i-k[i]层，最高不能超过n，最低不能小于1，给你一个起点和终点，问最少可以按几次到达目的地。在一个N层高的楼有一个奇怪的电梯，在每一层只能上升或下降一个特定的层数，中间不会停止，在给定的条件下，问能不能到达指定楼层，可以到达的话返回转操作次数，不可以的话返回-1.

附上代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#define M 205
#define MAX 0x3f3f3f3f
using namespace std;
int map[M][M],vis[M],dis[M];
int main()
{
int n,a,b,i,j,s;
while(~scanf("%d",&n)&&n)
{
scanf("%d%d",&a,&b);
memset(vis,0,sizeof(vis));
memset(dis,0,sizeof(dis));
for(i=1; i<=n; i++)
for(j=1; j<=n; j++)
{
if(i==j)
map[i][j]=0;  //同一个地方距离为0
else
map[i][j]=MAX;
}
for(i=1; i<=n; i++)
{
scanf("%d",&s);
if(i+s<=n)       //标记这一层电梯可以去的楼层
map[i][s+i]=1;
if(i-s>=1)
map[i][i-s]=1;
}
vis[a]=1;    //起点已走过
for(i=1; i<=n; i++)
dis[i]=map[a][i];   //初始化距离为每个点到起点的距离
int min,k,t;
for(i=1; i<=n; i++)
{
min=MAX;
for(j=1; j<=n; j++)
if(!vis[j]&&dis[j]<min)  //每次都找离终点最近的点
{
min=dis[j];
t=j;
}
vis[t]=1;               //标记为已经找过此点
for(j=1; j<=n; j++)
if(!vis[j]&&map[t][j]<MAX)   //从最近的点到下一个点的距离与初始距离进行比较
if(dis[j]>dis[t]+map[t][j])
dis[j]=dis[t]+map[t][j];
}
if(dis[b]<MAX)
printf("%d\n",dis[b]);
else           //不能到 则输出-1
printf("-1\n");
}
return 0;
}

邻接表代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define inf 0x3f3f3f3f
using namespace std;
struct Edge
{
int from,to,val,next;
}edge[40500];
int tol,s,t,n;
int dis[205];
bool vis[205];
int head[40500];

void init()
{
tol=0;
memset(head,-1,sizeof(head));
}

void addEdge(int u,int v)
{
edge[tol].from=u;
edge[tol].to=v;
edge[tol].val=1;
edge[tol].next=head[u];
head[u]=tol++;
}

void getmap()
{
int x;
for(int i=1;i<=n;i++)
{
scanf("%d",&x);
if(i-x>=1) addEdge(i,i-x);
if(i+x<=n) addEdge(i,i+x);
}
memset(vis,false,sizeof(vis));
memset(dis,inf,sizeof(dis));
}

void spfa()
{
queue<int>q;
q.push(s);
vis[s]=true;
dis[s]=0;
while(!q.empty())
{
int u=q.front();
q.pop();
vis[u]=false;
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if(dis[v]>dis[u]+edge[i].val)
{
dis[v]=dis[u]+edge[i].val;
if(!vis[v])
{
vis[v]=true;
q.push(v);
}
}
}
}
if(dis[t]<inf)
printf("%d\n",dis[t]);
else
printf("-1\n");
return;
}

int main()
{
int i,j;
while(~scanf("%d",&n)&&n)
{
scanf("%d%d",&s,&t);
init();
getmap();
spfa();
}
}