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POJ3050 Hopscotch 【DFS】

2024-08-01 10:39:12   219人阅读

Hopscotch
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 2113 Accepted: 1514

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 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.

Source

USACO 2005 November Bronze
暴搜，水题..

#include <stdio.h>
#include <string.h>
#include <map>

int G[5][5], ans;
const int mov[][2] = {0, 1, 0, -1, 1, 0, -1, 0};
std::map<int, int> mp;

void DFS(int x, int y, int k, int sum) {
    if(k == 6) {
        if(mp[sum]++ == 0) ++ans;
        return;
    }
    for(int i = 0, a, b; i < 4; ++i) {
        a = x+mov[i][0];
        b = y+mov[i][1];
        if(a>=0&&b>=0&&a<5&&b<5)
            DFS(a, b, k+1, sum*10+G[x][y]);
    }
}

int main() {
    int i, j;
    for(i = 0; i < 5; ++i)
        for(j = 0; j < 5; ++j)
            scanf("%d", &G[i][j]);
    for(i = 0; i < 5; ++i)
        for(j = 0; j < 5; ++j)
            DFS(i, j, 0, 0);
    printf("%d\n", ans);
    return 0;
}


POJ3050 Hopscotch 【DFS】


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