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 nn,
an n-dimensionaln−dimensional star
graph, also referred to as S_{n}S​n​​,
is an undirected graph consisting of n!n! nodes
(or vertices) and ((n-1)\
*\ n!)/2((n−1) ∗ 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}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}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-1n−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-1n−1 adjacent
nodes are obtained by swapping the first symbol and the d-thd−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, ..., nd=2,...,n.
For instance, in S_{4}S​4​​,
node 12341234 has 33 edges
connecting to nodes 21342134, 32143214,
and 42314231.
The following figure shows how S_{4}S​4​​ looks
(note that the symbols aa, bb, cc,
and dd 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:

nn:
the dimension of the star graph. We assume that nn ranges
from 44 to 99.
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

nn (dimension
of the star graph)

A list of 55 pairs
of nodes.

Output Format

A list of 55 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;

}

/*

*/