Hopscotch（dfs）

最新推荐文章于 2019-07-11 13:28:43 发布

sdau20163940

最新推荐文章于 2019-07-11 13:28:43 发布

阅读量233

点赞数

CC 4.0 BY-SA版权

分类专栏：搜索算法 2017年周训练赛

本文链接：https://blog.youkuaiyun.com/sinat_37668729/article/details/77071471

2017年周训练赛同时被 2 个专栏收录

73 篇文章

订阅专栏

搜索算法

37 篇文章

订阅专栏

本文探讨了Hopscotch游戏的算法实现，该算法通过在5x5数字网格上进行五次移动来创建六位数，并计算所有可能的不同六位数组合。采用深度优先搜索策略遍历所有路径并统计唯一结果。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Hopscotch

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4453		Accepted: 2973

Description

The cows play the child's game of hopscotch in a non-traditional way. Instead of a linear set of numbered boxes into which to hop, the cows create a 5x5 rectilinear grid of digits parallel to the x and y axes.

They then adroitly hop onto any digit in the grid and hop forward, backward, right, or left (never diagonally) to another digit in the grid. They hop again (same rules) to a digit (potentially a digit already visited).

With a total of five intra-grid hops, their hops create a six-digit integer (which might have leading zeroes like 000201).

Determine the count of the number of distinct integers that can be created in this manner.

Input

* Lines 1..5: The grid, five integers per line

Output

* Line 1: The number of distinct integers that can be constructed

Sample Input

Sample Output

Hint

OUTPUT DETAILS:
111111, 111112, 111121, 111211, 111212, 112111, 112121, 121111, 121112, 121211, 121212, 211111, 211121, 212111, and 212121 can be constructed. No other values are possible.

Source

USACO 2005 November Bronze

题意：

从任意一个点出发，任意走六步，共有多少种不同的数字组合方案。

思路：

搜索，遍历每一个点为起点时的情况，然后记录下来总的方案数，找不同的个数可以用个sort函数，使问题简单化。

代码：

#include <iostream>
#include <cstring>
#include <string>
#include <stdio.h>
#include <algorithm>
using namespace std;
int a[100010],mp[6][6],sum,ans;
int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
void dfs(int x,int y,int m,int n)   //xy代表坐标，m代表目前所形成的数，n代表长度
{
    int i;
    if(n==6)
    {
        a[ans++]=m;   //找到一种情况。
        return;
    }
    m=m*10+mp[x][y];
    for(i=0;i<4;i++)
    {
        int xx=x+dir[i][0];
        int yy=y+dir[i][1];
        if(xx>=0&&xx<5&&yy>=0&&yy<5)
        {
            dfs(xx,yy,m,n+1);
        }
    }
}
int main()
{
    int i,j;
    ans=0;
    for(i=0;i<5;i++)
        for(j=0;j<5;j++)
    {
        scanf("%d",&mp[i][j]);
    }
    for(i=0;i<5;i++)
        for(j=0;j<5;j++)
    {
        dfs(i,j,0,0);
    }
    sort(a,a+ans);
    sum=1;
    for(i=0;i<ans-1;i++)
    {
        if(a[i]!=a[i+1])sum++;
    }
    printf("%d\n",sum);
    return 0;
}