POJ3050 Hopscotch 【DFS】

最新推荐文章于 2024-11-22 22:31:58 发布

转载最新推荐文章于 2024-11-22 22:31:58 发布 · 54 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/blfshiye/p/4915587.html

本文介绍了一种基于Hopscotch游戏的算法实现，通过在一个5x5的数字网格中进行五次移动来生成六位数，并计算所有可能的不同整数的数量。提供了完整的C++代码实现，采用深度优先搜索策略解决此问题。

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

Sample Output

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;
}

转载于:https://www.cnblogs.com/blfshiye/p/4915587.html