Hopscotch(poj 3050)深度优先搜索

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本文探讨了一个基于5x5数字网格的游戏问题，通过不同路径跳跃形成六位数，使用DFS算法结合状态转移来计算所有可能形成的独特数字数量。

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来自《挑战程序设计竞赛》

1.题目原文

http://poj.org/problem?id=3050

Language:

Hopscotch

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3436		Accepted: 2368

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

2.解题思路

在5 * 5的方格里跳房子，起点是任意位置。将跳过的数连起来组成一个5位数（前导零可能），问一共能组成多少个数字？

当前的状态可以定义为当前位置、当前数字长度、当前数字这三个要素，利用状态转移dfs即可遍历所有情况。对于数字的储存使用set可以轻松满足要求。

3.AC代码

#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
#include<cmath>
using namespace std;

int Map[5][5];
int dx[]={1,-1,0,0};
int dy[]={0,0,1,-1};
set<int> res;

void dfs(int x,int y,int step,int number)
{
    if(step==5){
        res.insert(number);
        return;
    }
    else{
        for(int i=0;i<4;i++){
            int nx=x+dx[i];
            int ny=y+dy[i];
            if(0<=nx&&nx<5&&0<=ny&&ny<5){
                dfs(nx,ny,step+1,10*number+Map[nx][ny]);
            }
        }
    }
}

int main()
{
    for(int i=0;i<5;i++){
        for(int j=0;j<5;j++){
            cin>>Map[i][j];
        }
    }
    res.clear();
    for(int i=0;i<5;i++){
        for(int j=0;j<5;j++){
            dfs(i,j,0,Map[i][j]);
        }
    }
    cout<<res.size()<<endl;
    return 0;
}