hdu 1548 A strange lift（最短路）

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探讨了一种特殊电梯模型，电梯只能向上或向下跳跃特定楼层数，通过使用最短路径算法解决从任意起点到达终点所需的最少操作次数问题。

A strange lift

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19895 Accepted Submission(s): 7313

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 5
3 3 1 2 5
0

Sample Output

题意：n个点，每给点有一个权值v[i]，你可以选择往前跳v[i]或者往后跳v[i](不能<0或者>=n)问你从a到b最少跳几步，不能则输出-1

思路：最短路水题

题意：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;
#define N 210
#define INF 1000000000
int n,v[N];
int ma[N][N];
int d[N],vis[N];
void init()
{
    for(int i=0; i<n; i++)
        for(int j=0; j<n; j++)
            ma[i][j]=INF;
}
void djstl(int s)
{
    for(int i=0; i<n; i++)
    {
        d[i]= i==s?0:ma[s][i];
        vis[i]=(i==s);
    }
    for(int i=1; i<n; i++)
    {
        int maxn=INF;
        int v=-1;
        for(int j=0; j<n; j++)
            if(!vis[j]&&maxn>d[j])
            {
                maxn=d[j];
                v=j;
            }
        if(v==-1) continue;
        vis[v]=1;
        for(int j=0; j<n; j++)
            if(!vis[j]&&d[v]+ma[v][j]<d[j])
                d[j]=d[v]+ma[v][j];
    }
}
int main()
{
    int s,e;
    while(~scanf("%d",&n)&&n)
    {
        scanf("%d %d",&s,&e);
        for(int i=0; i<n; i++)
            scanf("%d",&v[i]);
        init();
        for(int i=0; i<n; i++)
        {
            if(i-v[i]>=0) ma[i][i-v[i]]=1;
            if(i+v[i]<n) ma[i][i+v[i]]=1;
        }
        djstl(s-1);
        if(d[e-1]==INF) printf("-1\n");
        else printf("%d\n",d[e-1]);

    }
    return 0;
}