hdu 3455 Leap Frog

最新推荐文章于 2020-09-12 09:46:06 发布

原创最新推荐文章于 2020-09-12 09:46:06 发布 · 705 阅读

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本文深入探讨了游戏开发领域的关键技术，包括游戏引擎、编程语言、硬件优化等，并重点介绍了AI音视频处理在游戏中的应用，如语音识别、图像处理、AR特效等。同时，文章还涉及了自动化测试、性能优化等现代开发流程，旨在为游戏开发者提供全面的技术指导。

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,…, x_n 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 x_n.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,…, x_n where 0 <=x₁,x₂,…, x_n<= 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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=101000;
int n;
int a[maxn];
int f[maxn][12];//dp[i][j]表示i位置,i+j位置的两个人
int main()
{
    while(scanf("%d",&n)==1&&n)
    {
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        memset(f,-1,sizeof(f));
        f[1][1]=0;
        for(int i=2;i<=n;i++)
        {
            for(int j=1;i+j<=n&&a[i+j]-a[i]<=10;j++)
            {
                for(int k=1;i-k>=1&&a[i+j]-a[i-k]<=10;k++)
                {
                    if(f[i-k][k]==-1) continue;
                    if(f[i][j]==-1) f[i][j]=f[i-k][k]+1;
                    else f[i][j]=min(f[i][j],f[i-k][k]+1);
                }
            }
        }
        int ans=-1;
        for(int i=n-1;i>=1&&a[n]-a[i]<=10;i--)
        {
            if(f[i][n-i]==-1) continue;
            if(ans==-1) ans=f[i][n-i];
            else ans=min(ans,f[i][n-i]);
        }
        printf("%d\n",ans);
    }
    return 0;
}