H - Hopscotch

最新推荐文章于 2025-05-29 11:38:29 发布

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最新推荐文章于 2025-05-29 11:38:29 发布

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本文链接：https://blog.youkuaiyun.com/huantingyizhan/article/details/104321643

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本文探讨了一种基于奶牛的非传统跳房子游戏，通过在一个5x5的数字网格中跳跃，创建六位数的整数。文章详细介绍了游戏规则，包括如何从任意格子开始，向上下左右四个方向跳跃，并在五次跳跃后形成一个六位数。通过深度优先搜索算法，计算了所有可能形成的独特整数数量。

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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.

#include <stdio.h>
#include <set>
#include <queue>
#include <cstring>
#include <algorithm>
using namespace std; 
int x1[4]={0,0,1,-1};
int y1[4]={1,-1,0,0};
set<int> st;//用于去重，自动排序， 常用st.insert();st.clear();st.find();st.size; 
int G[6][6];
bool text(int x,int y){//界限 
	if(x>=5||x<0||y>=5||y<0){
		return false;
	}
	return true;
}
void DFS(int x,int y,int k,int v){
	if(k==6){
		st.insert(v);
		return;
	}
	for(int i=0;i<4;i++){
		int newx=x+x1[i];
		int newy=y+y1[i];
		if(text(newx,newy)){
			k++;
			DFS(newx,newy,k,v*10+G[newx][newy]); 
			k--;
		}
	}
	
}
int main(){
	for(int i=0;i<5;i++){//绘图 
		for(int j=0;j<5;j++){
			scanf("%d",&G[i][j]);
		}
	}
	
	for(int i=0;i<5;i++){
		for(int j=0;j<5;j++){
			DFS(i,j,1,G[i][j]);
		}
	}
	printf("%d\n",st.size());
	return 0;
}