UVA 10422 Knights in FEN

本文介绍了一个基于5x5棋盘的黑白骑士问题,目标是最少步数内将初始棋盘状态转换为预定的目标状态。使用深度优先搜索算法解决该问题,并提供了一段C++代码实现。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

分析

dfs 回溯可解。为什么不剪枝?因为暂时没有什么剪枝的思路。

代码

#include <cstdio>

int G[5][5], startx, starty, min;
int A[5][5] = {1,1,1,1,1, 0,1,1,1,1, 0,0,-1,1,1, 0,0,0,0,1, 0,0,0,0,0};
int next[8][2] = {-2,-1, -2,1, -1,-2, -1,2, 1,-2, 1,2, 2,-1, 2,1};

void dfs(int x, int y, int s)
{
    if (s >= min) return;
    int n = 0;
    for (int i = 0; i < 5; i++)
        for (int j = 0; j < 5; j++)
            if (G[i][j] != A[i][j]) n++;
    if (n == 0) { min = s < min ? s: min; return; }

    for (int k = 0; k < 8; k++) {
        int nx = x + next[k][0];
        int ny = y + next[k][1];
        if (nx >= 0 && nx < 5 && ny >=0 && ny < 5) {
            int t = G[x][y];
            G[x][y] = G[nx][ny]; G[nx][ny] = t;
            dfs(nx, ny, s + 1);
            G[nx][ny] = G[x][y]; G[x][y] = t;
        }
    }
}

void solve()
{
    min = 11;
    dfs(startx, starty, 0);
    if (min == 11) printf("Unsolvable in less than 11 move(s).\n");
    else printf("Solvable in %d move(s).\n", min);
}

int main()
{
    int N;
    char c;
    scanf("%d\n", &N);
    while (N--) {
        for (int i = 0; i < 5; i++) {
            for (int j = 0; j < 5; j++) {
                if ((c = getchar()) == ' ') {
                    startx = i;
                    starty = j;
                    c = '0' - 1;
                }
                G[i][j] = c - '0';
            }
            getchar();
        }
        solve();
    }
    return 0;
}

题目

Description

There are black and white knights on a 5 by 5 chessboard. There are twelve of each color, and there is one square that is empty. At any time, a knight can move into an empty square as long as it moves like a knight in normal chess (what else did you expect?).

Given an initial position of the board, the question is: what is the minimum number of moves in which we can reach the final position which is:

Input

First line of the input file contains an integer N (N < 14) that indicates how many sets of inputs are there. The description of each set is given below:

Each set consists of five lines; each line represents one row of a chessboard. The positions occupied by white knights are marked by 0 and the positions occupied by black knights are marked by 1. The space corresponds to the empty square on board.

There is no blank line between the two sets of input.

Note: The first set of the sample input below corresponds to this configuration:

Output

For each set your task is to find the minimum number of moves leading from the starting input configuration to the final one. If that number is bigger than 10, then output one line stating

Unsolvable in less than 11 move(s).

otherwise output one line stating

Solvable in n move(s).

where n ≤ 10.

The output for each set is produced in a single line as shown in the sample output.

Sample Input

2
01011
110 1
01110
01010
00100
10110
01 11
10111
01001
00000

Sample Output

Unsolvable in less than 11 move(s).
Solvable in 7 move(s).
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值