POJ3050 Hopscotch

POJ3050 Hopscotch

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
2
3
4
5
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

1
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.

Answer

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#include <cstdio>  
#include <string>
#include <set>
using namespace std;

int maze[5][5];
set<string> s;
int dx[4] = { -1, 0, 0, 1 };
int dy[4] = { 0, -1, 1, 0 };

void dfs(int x, int y, string str);

int main() {
for (int i = 0; i < 5; ++i)
for (int j = 0; j < 5; ++j)
scanf("%d", &maze[i][j]);

for (int i = 0; i < 5; ++i)
for (int j = 0; j < 5; ++j)
dfs(i, j, "");

printf("%d\n", s.size());
return 0;
}

void dfs(int x, int y, string str) {
if(str.length() == 6) {
s.insert(str);
return;
}

for (int i = 0; i < 4; ++i) {
int nx = x + dx[i];
int ny = y + dy[i];
if (nx < 0 || nx > 4 || ny < 0 || ny > 4)
continue;
char ch = maze[nx][ny] + '0';
dfs(nx, ny, str + ch);
}
}

Analysis

  1. 在 5 * 5 的矩阵中,任一点出发,走出长度为 6 的序列,可以重复走,求出最后长度为 6 的不同序列的个数。
  2. dfs 遍历搜索。
  3. 每次序列存储为一个字符串,或者看成一个 6 位数,符合条件放入集合中去重即可。
# POJ

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