2017 ACM-ICPC 亚洲区（南宁赛区）网络赛 J Minimum Distance in a Star Graph 广度优先搜索

In this problem, we will define a graph called star graph, and the question is to find the minimum distance between two given nodes in the star graph.

Given an integer $n$, an $n-dimensional$ star graph, also referred to as S_{n}S​n​​, is an undirected graph consisting of $n!$ nodes (or vertices) and $((n-1)\ast n!)/2$ edges. Each node is uniquely assigned a label x_{1}\ x_{2}\ ...\ x_{n}x​1​​ x​2​​ ... x​n​​which is any permutation of the n digits $1,2,3,...,n$. For instance, an S_{4}S​4​​ has the following 24 nodes $1234,1243,1324,1342,1423,1432,2134,2143,2314,2341,2413,2431,3124,3142,3214,3241,3412,3421,4123,4132,4213,4231,4312,4321$. For each node with label x_{1}\ x_{2} x_{3}\ x_{4}\ ...\ x_{n}x​1​​ x​2​​x​3​​ x​4​​ ... x​n​​, it has $n-1$ edges connecting to nodes x_{2}\ x_{1}\ x_{3}\ x_{4}\ ...\ x_{n}x​2​​ x​1​​ x​3​​ x​4​​ ... x​n​​, x_{3}\ x_{2}\ x_{1}\ x_{4}\ ...\ x_{n}x​3​​ x​2​​ x​1​​ x​4​​ ... x​n​​, x_{4}\ x_{2}\ x_{3}\ x_{1}\ ...\ x_{n}x​4​​ x​2​​ x​3​​ x​1​​ ... x​n​​, ..., and x_{n}\ x_{2}\ x_{3}\ x_{4}\ ...\ x_{1}x​n​​ x​2​​ x​3​​ x​4​​ ... x​1​​. That is, the $n-1$ adjacent nodes are obtained by swapping the first symbol and the $d-th$ symbol of x_{1}\ x_{2}\ x_{3}\ x_{4}\ ...\ x_{n}x​1​​ x​2​​ x​3​​ x​4​​ ... x​n​​, for $d=2,...,n$. For instance, in S_{4}S​4​​, node $1234$ has $3$ edges connecting to nodes $2134$, $3214$, and $4231$. The following figure shows how S_{4}S​4​​ looks (note that the symbols $a$, $b$, $c$, and $d$ are not nodes; we only use them to show the connectivity between nodes; this is for the clarity of the figure).

In this problem, you are given the following inputs:

• $n$: the dimension of the star graph. We assume that $n$ ranges from $4$ to $9$.
• Two nodes x_{1}x​1​​ x_{2}x​2​​ x_{3}x​3​​ ... x_{n}x​n​​ and y_{1}y​1​​ y_{2}y​2​​ y_{3}\ ...\ y_{n}y​3​​ ... y​n​​ in S_{n}S​n​​.

You have to calculate the distance between these two nodes (which is an integer).

Input Format

$n$ (dimension of the star graph)

A list of $5$ pairs of nodes.

Output Format

A list of $5$ values, each representing the distance of a pair of nodes.

样例输入

4
1234 4231
1234 3124
2341 1324
3214 4213
3214 2143

样例输出

1
2
2
1
3

题目来源

2017 ACM-ICPC 亚洲区（南宁赛区）网络赛

题意：就是先给你一个数n，接下来有 n 行，每行输入两个数位为 n 的数 s 和 t ，问 s 怎么用最小的步骤变成 t ，变化条件：s 在每一步都能变成 n - 1 个数，假设 s 的每一个数位分别是：x1, x2, x3, ......, xn，则在一个步骤中， s 的第一个数位的数能跟后面的每一个数位交换一次，例如 s 是四位数x1x2x3x4，则 s 第一步能变成：x2x1x3x4 , x3x1x2x4 , x4x1x2x3，这三个数。

题解：这道题用广搜做。

附代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
#include <map>
using namespace std;
typedef long long LL;
///o(っ。Д。)っ AC万岁!!!!!!!!!!!!!!
const int maxn = 100;
struct node
{
    string book;
    int s;
}star, eq;
queue<node>first;


int bfs(int n)
{
    map<string, bool>maps;
    first.push(star);
    struct node exch, rec;
    while(!first.empty())
    {
        rec = first.front();
        first.pop();
        for(int i = 0; i < n; i++)
        {
            exch = rec;
            swap(exch.book[0], exch.book[i]);
            exch.s = rec.s + 1;
            if(exch.book == eq.book)
            {
                return exch.s;
            }
            if(!maps[exch.book])
            {
                first.push(exch);
                maps[exch.book] = true;  ///记得要保存每一次的变化，不然会爆内存
            }
        }
    }
}
int main()
{
    int n;
    while(scanf("%d ", &n) != EOF)
    {
        char ch;
        for(int i = 1; i <= 5; i++)
        {

            cin >> star.book;
            cin >> eq.book;
            star.s = eq.s = 0;
            printf("%d\n", bfs(n));

            while(!first.empty())
            {
                first.pop();
            }
        }
    }
    return 0;
}
/*


*/