Hopscotch

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.       

题意：在5*5的格子里，现在可以从任意点出发，上下左右移动，可以重复走过的点，每次只能走6步。问经过的格子中组成的6位数有多少种？

题解：遍历图中的每一个位置，然后从该位置开始暴力搜索，判断这个6位数之前出没出现过，记录走的步数，步数超过6步就停止

#include<iostream> #include<algorithm> #include<string.h> using namespace std; int path[6][6]; int ans; bool vis[10000005]; int dir_e[4][2]={{-1,0},{1,0},{0,-1},{0,1}}; void dfs(int x,int y,int keep,int n){ if(keep==6){ if(!vis[n]){ vis[n]=1; ans++; } return; } for(int i=0;i<4;i++){ int sx=x+dir_e[i][0]; int sy=y+dir_e[i][1]; if(sx>=0&&sy>=0&&sx<5&&sy<5) dfs(sx,sy,keep+1,n*10+path[sx][sy]); } } int main(){ for(int i=0;i<5;i++) for(int j=0;j<5;j++) cin>>path[i][j]; ans=0; memset(vis,0,sizeof(vis)); for(int i=0;i<5;i++) for(int j=0;j<5;j++) dfs(i,j,0,path[0][0]); dfs(0,0,0,path[0][0]); cout<<ans<<endl; return 0; }