2017-ACM南宁网络赛

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 nnn, an n−dimensionaln-dimensionaln−dimensional star graph, also referred to as SnS_{n}S​n​​, is an undirected graph consisting of n!n!n! nodes (or vertices) and ((n−1) ∗ n!)/2((n-1) * n!)/2((n−1) ∗ n!)/2 edges. Each node is uniquely assigned a label x1 x2 ... xnx_{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}1,2,3,...,n. For instance, an S4S_{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}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 x1 x2x3 x4 ... xnx_{1} x_{2} x_{3} x_{4} ... x_{n}x​1​​ x​2​​x​3​​ x​4​​ ... x​n​​, it has n−1n-1n−1 edges connecting to nodes x2 x1 x3 x4 ... xnx_{2} x_{1} x_{3} x_{4} ... x_{n}x​2​​ x​1​​ x​3​​ x​4​​ ... x​n​​, x3 x2 x1 x4 ... xnx_{3} x_{2} x_{1} x_{4} ... x_{n}x​3​​ x​2​​ x​1​​ x​4​​ ... x​n​​, x4 x2 x3 x1 ... xnx_{4} x_{2} x_{3} x_{1} ... x_{n}x​4​​ x​2​​ x​3​​ x​1​​ ... x​n​​, ..., and xn x2 x3 x4 ... x1x_{n} x_{2} x_{3} x_{4} ... x_{1}x​n​​ x​2​​ x​3​​ x​4​​ ... x​1​​. That is, the n−1n-1n−1 adjacent nodes are obtained by swapping the first symbol and the d−thd-thd−th symbol of x1 x2 x3 x4 ... xnx_{1} x_{2} x_{3} x_{4} ... x_{n}x​1​​ x​2​​ x​3​​ x​4​​ ... x​n​​, for d=2,...,nd = 2, ..., nd=2,...,n. For instance, in S4S_{4}S​4​​, node 123412341234 has 333 edges connecting to nodes 213421342134, 321432143214, and 423142314231. The following figure shows how S4S_{4}S​4​​ looks (note that the symbols aaa, bbb, ccc, and ddd 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:

• nnn: the dimension of the star graph. We assume that nnn ranges from 444 to 999.
• Two nodes x1x_{1}x​1​​ x2x_{2}x​2​​ x3x_{3}x​3​​ ... xnx_{n}x​n​​ and y1y_{1}y​1​​ y2y_{2}y​2​​ y3 ... yny_{3} ... y_{n}y​3​​ ... y​n​​ in SnS_{n}S​n​​.

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

Input Format

nnn (dimension of the star graph)

A list of 555 pairs of nodes.

Output Format

A list of 555 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

vis标记一次就够了

#include<cstdio>
#include<queue>
#include<map>
#include<string>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
string s,t,v;
int n;
struct node
{
    string s;
    int step;
    node(string str,int n)
    {
        s=str,step=n;
    }
    node()
    {
        s="";
        step=0;
    }
};
int bfs()
{
    queue<node>Q;
    map<string,bool>vis;
    Q.push(node(s,0));
    vis[s]=1;
    while(!Q.empty())
    {
        node u=Q.front();
        Q.pop();
        vis[u.s]=0;
        if(u.s==t)  return u.step;
        for(int i=1; i<n; ++i)
        {
            v=u.s;
            swap(v[0],v[i]);
            if(!vis[v])
            {
                node tc=node(v,u.step+1);
                vis[v]=1;
                Q.push(tc);
            }
        }
    }
}
int main()
{
    scanf("%d",&n);
    for(int i=1; i<=5; ++i)
    {
        cin>>s>>t;
        if(s==t)
        {
            puts("0");
            continue;
        }
        printf("%d
",bfs());
    }
}

#include<cstdio>
#include<queue>
#include<map>
#include<string>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
string s,t,u,v;
int n;
int bfs()
{
    map<string,int>mp;
    queue<string>Q;  //不能用queue<char*>;
    map<string,bool>vis;
    Q.push(s);
    mp[s]=0;
    vis[s]=1;
    while(!Q.empty())
    {
        u=Q.front();
        Q.pop();
        if(u==t)  return mp[t];
        for(int i=1; i<n; ++i)
        {
            v=u;
            swap(v[0],v[i]);
            mp[v]=mp[u]+1;
            if(!vis[v])
            {
                mp[v]=mp[u]+1;
                vis[v]=1;
                Q.push(v);
            }
        }
    }
}
int main()
{
    scanf("%d",&n);
    for(int i=1; i<=5; ++i)
    {
        cin>>s>>t;
        if(s==t)
        {
            puts("0");
            continue;
        }
        printf("%d
",bfs());
    }
}