2017 ACM-ICPC 亚洲区（南宁赛区）网络赛（J.Minimum Distance in a Star Graph）

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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

12213
题目大意：题目蛮长的，其实没什么内容，只看图和输入输出，这是与八数码类似的路径题，路径是每次可以有n-1种交换，第一个数与后面某位交换，照着之前那篇八数码的改了一下，就过了。。。
#include<iostream>#include<cstring>#include<cstdio>using namespace std;typedef int State[9];const int maxstate=1000000;State st[maxstate],goal;int dist[maxstate];int vis[362880],fact[9];int n;void init_lookup_table(){    fact[0]=1;    for(int i=1;i<=9;i++)        fact[i]=fact[i-1]*i;}int try_to_insert(int s){    int code =0;    for(int i=0;i<n;i++){        int cnt=0;        for(int j=i+1;j<n;j++) if(st[s][j]<st[s][i]) cnt++;        code +=fact[n-1-i]*cnt;    }    if(vis[code]) return 0;    return vis[code] =1;}int bfs(){    int front=1,rear=2;    while(front<rear){        State& s=st[front];        if(memcmp(goal,s,sizeof(s))==0) return front;        for(int d=1;d<n;d++)//第零位与后面的每一位一次交换        {            State& t=st[rear];            memcpy (&t,&s,sizeof(s));            t[d]=s[0];            t[0]=s[d];            dist[rear]=dist[front]+1;            if(try_to_insert(rear)) rear++;//判重与记录        }        front++;    }    return 0;}int main(){    init_lookup_table();    while(~scanf("%d",&n)){        char str[9];        for(int l=0;l<5;l++){            memset(st,0,sizeof(st));            memset(vis,0,sizeof(vis));            memset(dist,0,sizeof(dist));            scanf("%s",str);            for(int i=0;i<n;i++)                st[1][i]=int(str[i]-'1');            scanf("%s",str);            for(int i=0;i<n;i++)                goal[i]=int(str[i]-'1');            int ans=bfs();            printf("%d\n",dist[ans]);        }    }    return 0;}

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