hdu1548 A strange lift

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题目描述：

A strange lift

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17968    Accepted Submission(s): 6706

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 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 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".

 

Sample Input

5 1 53 3 1 2 50

 

Sample Output

3
大致题意：有一个n层的电梯，每层有一个数字k[i]，表示可以上升k[i]层，或者下降k[i]层。但是不能下降到小于1层，或者大于n层。给出两层a,b，问从第a层到第b层最少需要按多少次按钮？
分析：把每层看成有向图的顶点，如果从第i层可以一次到达第j层，则连一条从顶点i到顶点j的有向边，并且长度为1。转化为求有向图中顶点a到顶点b的最短路径长度。看到n最大为200，我先是使用了Floyd算法，然后TLE，后来使用Dijkstra算法。
超时代码：
#include<iostream>#include<stdio.h>#include<string.h>#include<cmath>using namespace std;const int N=300;int n,a,b,k[N],g[N][N];int main(){    while(~scanf("%d",&n))    {        if(n==0)break;        for(int i=1;i<=n;i++)            for(int j=1;j<=n;j++)            g[i][j]=10000;        cin>>a>>b;        for(int i=1;i<=n;i++)        {            scanf("%d",&k[i]);            if(i-k[i]>=1)                g[i][i-k[i]]=1;            if(i+k[i]<=n)                g[i][i+k[i]]=1;        }        for(int i=1;i<=n;i++)            for(int j=1;j<=n;j++)            for(int d=1;d<=n;d++)            g[i][j]=min(g[i][j],g[i][d]+g[d][j]);        cout<<g[a][b]<<endl;    }    return 0;}
AC代码：
#include<iostream>#include<stdio.h>#include<string.h>#include<cmath>using namespace std;const int N=300;const int MAX=100000000;int n,a,b,k[N],g[N][N];int vis[N],dist[N];void dijkstra(){    memset(vis,0,sizeof(vis));    for(int i=1;i<=n;i++)        dist[i]=g[a][i];    dist[a]=0;    vis[a]=1;    for(int i=1;i<=n;i++)    {        int mark=-1,mindis=MAX;        for(int j=1;j<=n;j++)        if(!vis[j]&&dist[j]<mindis)        {            mark=j;            mindis=dist[j];        }        if(mark==-1)            return;        vis[mark]=1;        for(int j=1;j<=n;j++)            if(!vis[j])            dist[j]=min(dist[j],dist[mark]+g[mark][j]);    }}int main(){    while(~scanf("%d",&n))    {        if(n==0)break;        for(int i=1;i<=n;i++)            for(int j=1;j<=n;j++)            g[i][j]=MAX;        cin>>a>>b;        for(int i=1;i<=n;i++)        {            scanf("%d",&k[i]);            if(i-k[i]>=1)                g[i][i-k[i]]=1;            if(i+k[i]<=n)                g[i][i+k[i]]=1;        }     dijkstra();     if(dist[b]==MAX)printf("-1\n");     else printf("%d\n",dist[b]);    }    return 0;}