最短路

最新推荐文章于 2021-03-11 00:06:20 发布

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本博客探讨了一个特殊电梯的问题，该电梯在每个楼层都有特定的上下行距离限制。给定起始楼层和目标楼层，作者通过构建图并应用Dijkstra算法，求解从起始楼层到达目标楼层所需的最少操作次数。对于无法达到目标楼层的情况，输出-1。通过实例展示了解决方法，并提供了完整的代码实现。

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A strange lift

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 16 Accepted Submission(s) : 5

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：给你一个起始点A和一个终止点B 让你求出最短的操作 从A到达B
思路：刚开始按照up和down的两种操作建图 然后用dijkstra()求解。
#include<stdio.h>
#define max 220
#define maxint 99999999
int map[max][max];
int vis[max];
int dis[max];
int num[max];
int N,A,B;
void init()
{
    int i,j;
    for(i=1; i<max; i++)
        for(j=1; j<max; j++)
        {
            map[i][j]=maxint;
            if(i==j)
                map[i][j]=0;
        }
}
void dijkstra()
{
    int i,j;
    for(i=1; i<=N; i++)
    {
        vis[i]=0;
        dis[i]=map[A][i];
    }
    vis[i]=1;
    for(i=1; i<N; i++)
    {
        int mid=0;
        int d=maxint;
        for(j=1; j<=N; j++)
        {
            if(vis[j]==0&&dis[j]<d)
            {
                mid=j;
                d=dis[j];
            }
        }
        vis[mid]=1;
        if(d==maxint)
            break;
        for(j=1; j<=N; j++)
        {
            if(vis[j]==0&&map[mid][j]<=B&&map[mid][j]+d<dis[j])
                dis[j]=map[mid][j]+d;
        }
    }
}
int main()
{
    while(scanf("%d",&N),N)
    {
        scanf("%d%d",&A,&B);
        init();
        for(int i=1; i<=N; i++)
        {
            scanf("%d",&num[i]);
            if(i+num[i]<=B)
                map[i][i+num[i]]=1;
            if(i-num[i]>=1)
                map[i][i-num[i]]=1;
        }
        dijkstra();
        if(dis[B]==maxint)
            printf("-1\n");
        else
            printf("%d\n",dis[B]);
    }
    return 0;
}