UVA 1533 - Moving Pegs 状态压缩 + bfs

本文链接：https://blog.youkuaiyun.com/s_h_r/article/details/45157827

本文介绍了一种解决特殊棋盘问题的算法实现，棋盘上每个位置可以为空或者放置一个球。球只能通过相邻且有球的位置跳跃，并且中间必须隔着一个空位。文章详细展示了如何使用广度优先搜索(BFS)来寻找从初始状态达到所有球都占据指定位置所需的最少步骤。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

这题需要注意的是当一个球的相邻位置为空时不能直接跳过去之间必须要有球垫着

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

#define bug puts("bug")

const int link[15][6] = { {-1,-1,-1,-1, 1, 2},  {-1, 0,-1, 2, 3, 4},  { 0,-1, 1,-1, 4, 5},  {-1, 1,-1, 4, 6, 7},
                                    { 1, 2, 3, 5, 7, 8},  { 2,-1, 4,-1, 8, 9},  {-1, 3,-1, 7,10,11},  { 3, 4, 6, 8,11,12},
                                    { 4, 5, 7, 9,12,13},  { 5,-1, 8,-1,13,14},  {-1, 6,-1,11,-1,-1},  { 6, 7,10,12,-1,-1},
                                    { 7, 8,11,13,-1,-1},  { 8, 9,12,14,-1,-1},  { 9,-1,13,-1,-1,-1} };

struct Node{
    int g;   // 0 represent empty and 1 represent full
    int step;
    int num;
    Node(int g = 0, int step = 0, int num = 0) : g(g), step(step), num(num){}
};

struct Path{
    int u, v;
    Path(int u = 0, int v = 0) : u(u), v(v){}
};

const int maxn = 16;

int n, cnt, k;
int fa[1 << maxn];
bool vis[1 << maxn];
Path path[1 << maxn];
Node S;

void print(int c){
    if(fa[c] == -1) return;
    print(fa[c]);
    if(c == k) printf("%d %d\n", path[c].u + 1, path[c].v + 1);
    else printf("%d %d ", path[c].u + 1, path[c].v + 1);
}

int bfs(){
    queue<Node> Q;
    Q.push(S);
    vis[Q.front().g] = true;
    while(!Q.empty()){
//            bug;
        Node cur = Q.front(); Q.pop();
        if(cur.g == (1 << n)){
            k = cur.num;
            return cur.step;
        }
        for(int i = 0; i < 15; i++)if((1 << i) & cur.g){ //if full

            for(int j = 0; j < 6; j++){
                int x = (1 << i);
                int st = 0;
                int go = link[i][j];
                while(go != -1){ // find empty
                    if(!((1 << go) & cur.g)){
                        if(!st) break;
                        Node tmp;
                        x ^= cur.g;
                        x |= (1 << go);
                        tmp.g = x;
                        tmp.step = cur.step+1;

                        if(!vis[x]){
                            vis[x] = true;
                            tmp.num = ++cnt;
                            fa[cnt] = cur.num;
                            path[cnt] = Path(i, go);
                            Q.push(tmp);

                        }
                        break;
                    }
                    x |= (1 << go); // save
                    go = link[go][j];
                    st++;
                }
            }
        }
    }
    return -1;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        memset(vis, false, sizeof(vis));
        n--;
        int g = 0;
        for(int i = 0; i < 15; i++){
            if(i != n)
                g |= (1 << i);
        }
        S.g = g;
        S.step = 0;
        S.num = cnt = 0;
        fa[0] = -1;
        int ans = bfs();

        if(ans == -1){
            printf("IMPOSSIBLE\n");
        }
        else{
           printf("%d\n", ans);
           print(k);
        }

    }
    return 0;
}