POJ--3050--BFS

最新推荐文章于 2021-04-25 18:36:38 发布

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最新推荐文章于 2021-04-25 18:36:38 发布

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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
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 2 1
1 1 1 1 1
Sample Output
15
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：常规BFS,需要注意的是vis[]数组空间尽可能大，但是二维vis[][]就不存在这样的情况

#include<iostream>
#include<cstring>
using namespace std;
int a[6][6],vis[10000000];
int ans;
//int dx[]={1,0,-1,0},dy[]={0,1,0,-1};
//int dx[]={1,-1,0,0},dy[]={0,0,1,-1};
int dx[]={0,0,-1,1},dy[]={-1,1,0,0};//三种方法都可以 
void dfs(int x,int y,int len,int cur){
	if(len==6){//此时作为出口，用ans统计当前点 
		if(!vis[cur]){ 
			vis[cur]=1;
			ans++; 
		}
		return;
	}
	for(int i=0;i<4;i++){
		int nx=x+dx[i];
		int ny=y+dy[i];
		if(nx>=0&&ny>=0&&nx<5&&ny<5)
			dfs(nx,ny,len+1,cur*10+a[nx][ny]);//这里由于需要*10，因此数组vis[]需要开辟足够大 
	}
	return;		
}
int main(){
	ans=0;
	memset(vis,0,sizeof(vis));
	for(int i=0;i<5;i++)
		for(int j=0;j<5;j++)
			cin>>a[i][j];
	for(int i=0;i<5;i++)
		for(int j=0;j<5;j++)
			dfs(i,j,1,a[i][j]);
	cout<<ans<<endl;
}