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

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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 $n - d i m e n s i o n a l$ star graph, also referred to as $S_{n}$ , is an undirected graph consisting of $n!$ nodes (or vertices) and $((n-1)\ *\ n!)/2$ edges. Each node is uniquely assigned a label $x_{1}\ x_{2}\ ...\ x_{n}$ which is any permutation of the n digits $1, 2, 3, . . ., n$ . For instance, an $S_{4}$ 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 $x_{1}\ x_{2} x_{3}\ x_{4}\ ...\ x_{n}$ , it has $n - 1$ edges connecting to nodes $x_{2}\ x_{1}\ x_{3}\ x_{4}\ ...\ x_{n}$ , $x_{3}\ x_{2}\ x_{1}\ x_{4}\ ...\ x_{n}$ , $x_{4}\ x_{2}\ x_{3}\ x_{1}\ ...\ x_{n}$ , ..., and $x_{n}\ x_{2}\ x_{3}\ x_{4}\ ...\ x_{1}$ . That is, the $n - 1$ adjacent nodes are obtained by swapping the first symbol and the $d - t h$ symbol of $x_{1}\ x_{2}\ x_{3}\ x_{4}\ ...\ x_{n}$ , for $d = 2, . . ., n$ . For instance, in $S_{4}$ , node $1234$ has $3$ edges connecting to nodes $2134$ , $3214$ , and $4231$ . The following figure shows how $S_{4}$ 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 $x_{1}$ $x_{2}$ $x_{3}$ ... $x_{n}$ and $y_{1}$ $y_{2}$ $y_{3}\ ...\ y_{n}$ in $S_{n}$ .

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

41234 42311234 31242341 13243214 42133214 2143

样例输出

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;
}
/*

*/

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