hdu 1548 A strange lift

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本文介绍了一种特殊电梯的工作原理，电梯只能上行或下行特定楼层数。通过使用Dijkstra算法，解决从任意起点到达终点所需的最少按钮按压次数的问题。

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A strange lift

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28013 Accepted Submission(s): 10050

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

Sample Output

做了好多题了，终于发现技巧了，可惜给弄成双向了。

Dijkstra：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
int n, a, b;
int s[210];
int cost[210][210];
int dp[210];
void Dijkstra()
{
    int vis[210] = {0};
    int i, j, pos, mi;
    for(i = 1; i <= n; i++)
    {
        dp[i] = cost[a][i];
    }
    vis[a] = 1;
    dp[a] = 0;
    for(i = 1; i <= n; i++)
    {
        mi = inf;
        for(j = 1; j <= n; j++)
        {
            if(dp[j] < mi && !vis[j])
            {
                mi = dp[j];
                pos = j;
            }
        }
        if(mi == inf)
            break;
        vis[pos] = 1;
        for(j = 1; j <= n; j++)
        {
            if(dp[j] > dp[pos] + cost[pos][j] && !vis[j])
                dp[j] = dp[pos] + cost[pos][j];
        }
    }
}
int main()
{
    while(cin >> n, n)
    {
        cin >> a >> b;
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n; j++)
                cost[i][j] = inf;
            cost[i][i] = 0;
        }
        for(int i = 1; i <= n; i++)
        {
            cin >> s[i];
            if(i - s[i] >= 1 && i - s[i] <= n)
                cost[i][i - s[i]] = 1;//千万别写成cost[i][i-s[i]]=cost[i-s[i]][i]=1;
            if(i + s[i] >= 1 && i + s[i] <= n)
                cost[i][i + s[i]] = 1;//这也是重点
        }
        Dijkstra();
        if(dp[b] == inf)
            printf("-1\n");
        else
            printf("%d\n", dp[b]);
        // for(int i=1;i<=n;i++)
        // {
        // 	for(int j=1;j<=n;j++)
        // 		printf("%d ",cost[i][j]);
        // 	printf("\n");
        // }
    }
    return 0;
}