2017ACM-ICPC南宁网络赛Frequent Subsets Problem（康托展开+bfs）

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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 $n!)/2((n-1)\ *\ n!)/2$ 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：给定两个序列，只能用第一个和后面的交换，能把第一个序列换成第二个序列，求最小的交换次数.

我们队没有看出来只能第一个和最后一个交换，然后就感觉这题没法做，赛后他们说只能第一个和后面的交换，*********这*****不就是康托展开+bfs么

不会康托展开请戳这里

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <queue>#include <algorithm>using namespace std;struct node{int ct;char s[99];int step;};char s1[99], s2[99];int a[99];int v[363880];int f(int n){if(n == 0)return 1;int i = 1;int x = 1;for(i = 1;i <= n;i++)x *= i;return x;}int cantor(char s[], int n){int i, j, sum;for(i = 0;i < n;i++){sum = 0;for(j = i + 1;j < n;j++){if(s[j] < s[i])sum++;}a[n-i-1] = sum;}int nx = 0;for(i = 0;i < n;i++){nx = nx + a[i] * f(i);}return nx;}void bfs(int n, int nx){memset(v,0,sizeof(v));node a, b;queue<node>q;strcpy(a.s,s1);a.ct = cantor(a.s,n);a.step = 0;v[a.ct] = 1;q.push(a);while(!q.empty()){a = q.front();q.pop();if(a.ct == nx){cout<<a.step<<endl;return;}b.step = a.step + 1;for(int i = 1;i < n;i++){strcpy(b.s,a.s);swap(b.s[0],b.s[i]);b.ct = cantor(b.s,n);if(!v[b.ct]){q.push(b);v[b.ct] = 1;}}}}int main(){int n, sum;int nx;cin>>n;for(int k = 0;k < 5;k++){cin>>s1>>s2;nx = cantor(s2,n);bfs(n,nx);}return 0;}

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