HDU 3455 Leap Frog（线性DP)

Problem Description

Jack and Jill play a game called "Leap Frog" in which they alternate turns jumping over each other. Both Jack and Jill can jump a maximum horizontal distance of 10 units in any single jump. You are given a list of valid positions x1,x2,…,
xn where Jack or Jill may stand. Jill initially starts at position x1, Jack initially starts at position x2, and their goal is to reach position xn.Determine the minimum number of jumps needed until either Jack or
Jill reaches the goal. The two players are never allowed to stand at the same position at the same time, and for each jump, the player in the rear must hop over the player in the front.



Input

The input file will contain multiple test cases. Each test case will begin with a single line containing a single integer n (where 2 <= n <= 100000). The next line will contain a list of integers x1,x2,…, xn where 0 <=x1,x2,…,
xn<= 1000000. The end-of-fi le is denoted by a single line containing "0".



Output

For each input test case, print the minimum total number of jumps needed for both players such that either Jack or Jill reaches the destination, or -1 if neither can reach the destination.



Sample Input

6
3 5 9 12 15 17
6
3 5 9 12 30 40

Sample Output

3
-1
用DP[i][j]表示：第一个人到了I点距离第二个人j的最小步数。dp[i][j]=min{dp[j][k]+1}.
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
typedef long long LL;
using namespace std;
const int INF=0x3f3f3f;
const int maxn=1e5+100;
int dp[maxn][15];
int hash[10*maxn];
int num[maxn],n;
int main()
{
    while(~scanf("%d",&n)&&n)
    {
        memset(dp,INF,sizeof(dp));
        memset(hash,-1,sizeof(hash));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&num[i]);
            hash[num[i]]=i;
        }
        dp[2][num[2]-num[1]]=0;
        for(int i=2;i<=n;i++)
        {
            for(int j=1;j<=10;j++)
            {
                if(j>num[i])  break;
                for(int k=j+1;k<=10;k++)
                {
                    int tt=num[i]-j;
                    if(hash[tt]>=0)
                        dp[i][j]=min(dp[hash[tt]][k-j]+1,dp[i][j]);//距离小于10的前面点中递推
                }
            }
        }
        int ans=INF;
        for(int i=1;i<=10;i++)
//        {
//            cout<<"1111   "<<dp
[i]<<endl;
            ans=min(ans,dp
[i]);
//        }
        if(ans<INF) printf("%d\n",ans);
        else   printf("-1\n");
    }
    return 0;
}