poj 3050 Hopscotch（dfs暴力）

Hopscotch

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3951 Accepted: 2627

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

1 1 1 1 11 1 1 1 11 1 1 1 11 1 1 2 11 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：给出一个5*5的矩阵，要求在矩阵中走6步，组成一个6位数字，问有多少种数字。

分析：用dfs对每一个点开始，然后搜索一遍，用vis标志这个数字有没有出现过，统计一下即可。

代码如下：

#include <stdio.h>#include <string.h>int map[20][20];int vis[1000000];int dir[4][2]={-1,0,1,0,0,-1,0,1};int cnt=0;void dfs(int x,int y,int val,int t){//printf("<%d %d %d %d>\n",x,y,val,t);if(t==5){if(vis[val]==0){vis[val]=1;cnt++;}return ;}for(int k=0;k<4;k++){int px=x+dir[k][0];int py=y+dir[k][1];if(px>=0 && px<5 && py>=0 && py<5)dfs(px,py,val*10+map[px][py],t+1);}}int main(){int i,j;for(i=0;i<5;i++)for(j=0;j<5;j++)scanf("%d",map[i]+j);memset(vis,0,sizeof(vis));for(i=0;i<5;i++){for(j=0;j<5;j++){dfs(i,j,map[i][j],0);}}printf("%d\n",cnt);return 0;}