AcWing 1107. 魔板（搜索）

题目链接

题目描述

• 见题目链接！

解题思路

1. 将每一种魔板看作一个点，从初始状态12345678搜到目标状态
2. 用unordered_map进行哈希，如：<String, int> ==> <"21345678", 2>，指从初始状态到目前状态需要操作多少次
3. 因为还要输出进行的操作序列，如<String, pair<char, String>> ==> <"X", <'A', Y>>，指从Y进行操作A得到X

题目代码

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

using namespace std;

char g[2][4];
unordered_map<string, pair<char, string>> pre;
unordered_map<string, int> dist;

void set(string state)
{
    for (int i = 0; i < 4; i ++ ) g[0][i] = state[i];
    for (int i = 7, j = 0; j < 4; i --, j ++ ) g[1][j] = state[i];
}

string get()
{
    string res;
    for (int i = 0; i < 4; i ++ ) res += g[0][i];
    for (int i = 3; i >= 0; i -- ) res += g[1][i];
    return res;
}

string move0(string state)
{
    set(state);
    for (int i = 0; i < 4; i ++ ) swap(g[0][i], g[1][i]);
    return get();
}

string move1(string state)
{
    set(state);
    int v0 = g[0][3], v1 = g[1][3];
    for (int i = 3; i >= 0; i -- )
    {
        g[0][i] = g[0][i - 1];
        g[1][i] = g[1][i - 1];
    }
    g[0][0] = v0, g[1][0] = v1;
    return get();
}

string move2(string state)
{
    set(state);
    int v = g[0][1];
    g[0][1] = g[1][1];
    g[1][1] = g[1][2];
    g[1][2] = g[0][2];
    g[0][2] = v;
    return get();
}

int bfs(string start, string end)
{
    if(start == end) return 0;
    
    queue<string> q;
    q.push(start);
    dist[start] = 0;
    
    while(!q.empty())  // queue.empty()
    {
        auto t = q.front();  // queue.front()
        q.pop();  // queue.pop()
        
        string m[3];
        m[0] = move0(t);
        m[1] = move1(t);
        m[2] = move2(t);
        
        for(int i = 0; i < 3; i ++ )
            if(!dist.count(m[i]))  // unordered_map.count()
            {
                dist[m[i]] = dist[t] + 1;
                pre[m[i]] = {'A' + i, t};
                q.push(m[i]);  // queue.push()
                if(m[i] == end) return dist[end];
            }
    }
    
    return -1;
}

int main()
{
    int x;
    string start, end;
    for(int i = 0; i < 8; i ++ )
    {
        cin >> x;
        end += char(x + '0');
    }
    for(int i = 1; i <= 8; i ++ ) start += char(i + '0');
    
    int step = bfs(start, end);
    
    cout << step << endl;
    
    string res;
    while(end != start)
    {
        res += pre[end].first;
        end = pre[end].second;
    }
    
    reverse(res.begin(), res.end());  // string.revrse()
    
    if(step > 0) cout << res << endl;
    
    return 0;
}

标签：魔板,return,string,int,end,1107,state,dist,AcWing
来源： https://www.cnblogs.com/esico/p/16474659.html