HDU 1548 A strange lift 最短路 题解

HDU 1548 A strange lift

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

3

分析：

数列K的每项代表了每层只能下降或上升的层数，能到达的楼层为1~N之间。每次我们找s到哪一层需要的次数最少，那么这一层已经最优化了，不能间接通过其他楼层来优化自己的路径次数，但其他楼层也许可以借助这一层来优化自己的路径次数。经过循环判断一个个的找到s到每层的最短路径次数，但有些楼层是不能到达的，最差经过N次后要么找到可以到达情况下的最短路径次数、要么无法发到达。

代码：

#include <stdio.h>
#include <string.h>

int MAX = 0xafafafa;
int n;
int Map[205][205];//map[i][j]表示从i点到j点可以一次到达；
int a[205],cnt;
int vis[205],cast[205];//vis[]数组用来标记是否找到最短路，cast[]用来存储起点s到1~n各点的当前最短路径长度；

void zui_duan_lu(int s,int e)
{
    int i,j,Min,pos;
    memset(vis,0,sizeof(vis));//设s到所有点的最终最短路都还没找到；
    for(i = 0; i<n; i++)
        cast[i] = Map[s][i];//初始化cast[]的当前最短路径
    cast[s] = 0;//s-->s的最短路径为0；
    vis[s] = 1;//s-->s的最短路径已找到；
    for(i = 1; i<n; i++)
    {
        Min = MAX;
        /*遍历搜索当前s到哪一点的路径最短(最短的记为pos点)，那么该点(pos)的最终最短路已经找到，
        因为该点已经无法间接通过其他点来减少s到其的最短路径*/
        for(j = 0; j<n; j++)
        {
            if(cast[j]<Min && !vis[j])
            {
                pos = j;//最小点的下标
                Min = cast[j];//最小点的最短路径长度
            }
        }
        if(Min == MAX)//vis[]所有点都已经找到最终最短路径，结束
            break;
        vis[pos] = 1;//标记找到的s-->pos最终最短路径
        for(j = 0; j<n; j++)//遍历搜索其他点(j点)借助pos点是否能减少到s点的最短路径长度
        {
            if(cast[pos]+Map[pos][j]<cast[j] && !vis[j])//该点能优化到达s点的最短路径且该点未找到最终最短路径
                cast[j] = cast[pos]+Map[pos][j];//更新该点到s点的最短路径长度
        }
    }
}

int main()
{
    int i,j,s,e,x,y;
    while(~scanf("%d",&n),n)
    {
        scanf("%d%d",&s,&e);
        for(i = 0; i<n; i++)
            for(j = 0; j<n; j++)
                Map[i][j] = MAX;//初始化Map[]
        for(i = 0; i<n; i++)
        {
            scanf("%d",&a[i]);
            if(i+a[i]<n)
                Map[i][i+a[i]] = 1;//i-->i+a[i]可以一步到达
            if(i-a[i]>=0)
                Map[i][i-a[i]] = 1;//i-->i-a[i]可以一步到达
        }
        zui_duan_lu(s-1,e-1);
        printf("%d\n",cast[e-1]==MAX?-1:cast[e-1]);
    }

    return 0;
}