Hopscotch POJ - 3050

最新推荐文章于 2020-05-07 11:19:02 发布

原创最新推荐文章于 2020-05-07 11:19:02 发布 · 182 阅读

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奶牛以独特的方式玩跳房子游戏，通过在一个5x5的矩形网格中跳跃数字，形成六位整数。通过五次网格内的跳跃，它们创建了一个可能带有前导零的六位数。问题是要找出所有可以通过这种方式构建的不同整数的个数。输入包含一个5x5的数字网格，输出为可构建的不同整数的数量。提示指出，可以通过DFS搜索每个位置的四个方向（只要不越界），并使用集合来存储不重复的步数。

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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.

思路：最直接对每一个位置进行DFS，但是需要注意的是可以重复走，所以每次遍历时都是四个方向，只要不越界就可以；用set不重复的特性来记录总的步数

#include<iostream>
#include<cstdio>
#include<fstream>
#include<queue>
#include<vector>
#include<algorithm>
#include<cmath>
#include<set>
#include<cstring>
#define MAX 110000
#define INF 10000000

using namespace std;

int a[5][5];
int dix[4] = {1,-1,0,0};
int diy[4] = {0,0,-1,1};
set <int> hop;
int ans = 0; 

//DFS 
void Dfs(int x, int y, int pos, int num){
	//返回的唯一条件是走了五步,返回时判断路径是否首次经过 
	if(pos == 5){
		hop.insert(num);
		return ;
	}	
	//遍历
	for(int i=0; i<4; i++){
		int nx = x + dix[i];
		int ny = y + diy[i];
		if(nx>=0 && ny>=0 && nx<5 && ny <5){
			Dfs(nx, ny, pos+1, num*10+a[nx][ny]);
		}
	} 
}

//处理 
void Solve(){
	for(int i=0; i<5; i++)
		for(int j=0; j<5; j++)
			Dfs(i,j,0,a[i][j]);
	cout << hop.size() << endl;
}

//测试函数 
int main(){
	ifstream in ("D:\\钢铁程序员\\程序数据\\034HopscotchPOJ3050.txt");//从文件读取数据流，省去手动输入的麻烦 
	if(!in){//读取如果失败 
		cout << "ERROR" << endl;
	}
	for(int i=0; i<5; i++)
		for(int j=0; j<5; j++)
			in >> a[i][j];
	Solve(); 
	in.close();//打开文件以后要关闭 
	return 0;
}