Minimum Distance in a Star Graph（南宁网络赛）

来源：互联网发布：吴江法院拍卖公告淘宝编辑：程序博客网时间：2024/06/11 09:51

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

//题意：给定一个n，然后有 n! 个顶点，顶点的真实值分别为123...n，123...(n-1)n，... 举个例子：n=4，顶点真实值分别为

$1234, 1243, 1324, 1342, 1423, 1432, 2134, 2143, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 3412, 3421, 4123, 4132, 4213, 4231, 4312, 4321$

每个顶点仅与另外n-1个点之间有边，每个点与和它相连的点是有规律的（后面解释），且边权值都为 1 。仍举n=4的例子，1234与2134，3214，4231相连，2134与1234，3124，4132相连，abcd与bacd，cbad，dbca相连，就是这个顶点真实值的第一位数分别与它后面的各位数交换。

现在给你2个顶点的真实值，让你计算这两点间的最短路径。（仅5组）

//思路：写两个函数，一个是把顶点的真实值转换为序号，一个是把序号转换为顶点的真实值。比如：1234 -> 1，2134 -> 7；5 -> 1423， 8 -> 2143 ... 顶点的真实值越小，它的序号就越小，分别为1、2、3、... n! 。然后就可以把题目中描述的关系建一张图（9! =362880 每个点最多连8条边，二维数组、vector都可以存），然后跑SPFA即可。

本人水平有限，这两个函数写的巨长...，写的自己都有点晕了...，还好RP够高，1A了，hhh

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <cmath>#include <vector>#include <queue>#include <algorithm>using namespace std;const int MAX=362880+10;const int INF=1e9;int n;int num;vector<int>map[MAX];//把真实值转换为序号int change(int x){    int temp=x;    int res=num;    int arr[100];    int ans[100];    int tot=0;    for(int i=0;;i++)    {        if(temp==0)            break;        arr[i]=temp%10;        temp=temp/10;        tot++;    }    for(int i=1;i<tot;i++)        ans[i]=i;    int sum=0;    for(int i=n;i>0;i--)    {        res=res/i;        int tmp=n-i+1;        int w=1;        for(int j=1;j<tot;j++)        {            if(ans[j]==-1)                continue;            if(ans[j]==arr[tot-tmp])                break;            w++;        }        sum+=res*(w-1);        ans[arr[tot-tmp]]=-1;    }    return sum+1;}//把序号转换为真实值int trans(int x){    int temp=x;    int arr[100];    int ans[100];    int cnt=1;    int res=num;    for(int i=1;i<=n;i++)        ans[i]=i;    for(int i=n;i>0;i--)    {        res=res/i;        int w=1;        int tt=temp/res;        if(temp%res!=0)            tt++;        for(int j=1;j<=n;j++)        {            if(ans[j]==-1)            {                continue;            }            if(w==tt)            {                w=j;                break;            }            w++;        }        arr[cnt]=ans[w];        ans[w]=-1;        cnt++;        temp=temp-res*(temp/res);        if(temp==1||temp==0)            break;    }    if(temp==1)    {        for(int i=1;i<=n;i++)        {            if(ans[i]==-1)                continue;            arr[cnt++]=i;        }    }    else if(temp==0)    {        for(int i=n;i>=1;i--)        {            if(ans[i]==-1)                continue;            arr[cnt++]=i;        }    }    int sum=0;    for(int i=1;i<cnt;i++)    {        sum+=arr[i]*pow(10,n-i);    }    return sum;}//最短路径算法SPFAint dis[MAX];void SPFA(int x){    for(int i=1;i<=num;i++)        dis[i]=INF;    queue<int>q;    q.push(x);    dis[x]=0;    while(!q.empty())    {        int now=q.front();        q.pop();        for(int i=0;i<map[now].size();i++)        {            int v=map[now][i];            if(dis[v]>dis[now]+1)            {                dis[v]=dis[now]+1;                q.push(v);            }        }    }}int main(){    scanf("%d",&n);    num=1;    for(int i=2;i<=n;i++)        num*=i;             //建图       for(int i=1;i<=num;i++)    {        int xnum=trans(i);        int temp=xnum;        int arr[100];        int cnt=1;        int sum;        for(int j=1;;j++)        {            arr[cnt++]=temp%10;            temp=temp/10;            if(temp==0)                break;        }        temp=arr[n];        for(int j=1;j<n;j++)        {            sum=0;            arr[n]=arr[j];            arr[j]=temp;            for(int k=1;k<=n;k++)            {                sum+=arr[k]*pow(10,k-1);            }            sum=change(sum);            //无向边            map[i].push_back(sum);            map[sum].push_back(i);            arr[j]=arr[n];            arr[n]=temp;        }    }        int T=5;    while(T--)    {        int x,y;        scanf("%d%d",&x,&y);        x=change(x);        y=change(y);        SPFA(x);        printf("%d\n",dis[y]);    }    return 0;}

阅读全文

0 0