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

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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

题解：

题意：

给一个全排序，每次允许第一个数字和后面的随便一个数字交换，让你求可以达到目标串的最小交换次数

题解：

一开始从后往前放wa了几发，后来发现9的阶乘是36w左右，满足bfs的时间复杂度，然后就bfs+状态判断ac了

代码：

#include<iostream>#include<cstring>#include<stdio.h>#include<math.h>#include<string>#include<stdio.h>#include<queue>#include<stack>#include<map>#include<vector>#include<deque>#include<algorithm>#define ll long long#define INF 1008611111#define M (t[k].l+t[k].r)/2#define lson k*2#define rson k*2+1using namespace std;struct node{    ll v;    int step;};queue<node>q;map<ll,int>p;int a[15];int b[15];int main(){    int test,n,i,j;    ll tar,cur;    node now,next;    scanf("%d",&n);    test=5;    while(test--)    {        scanf("%lld%lld",&tar,&cur);        p.clear();        if(tar==cur)        {            printf("0\n");            continue;        }        while(!q.empty())            q.pop();        now.v=cur;        now.step=0;        p[now.v]=1;        q.push(now);        int t;        while(!q.empty())        {            now=q.front();            q.pop();            for(i=0;i<n;i++)            {                a[n-i-1]=now.v%10;                b[n-i-1]=a[n-i-1];                now.v/=10;            }            for(i=1;i<n;i++)            {                swap(b[0],b[i]);                next.v=0;                for(j=0;j<n;j++)                {                    next.v=next.v*10+b[j];                    b[j]=a[j];                }                if(!p[next.v])                {                    p[next.v]=1;                    next.step=now.step+1;                    if(next.v==tar)                    {                        t=next.step;                        goto loop;                    }                    q.push(next);                }            }        }        loop:;        printf("%d\n",t);    }    return 0;}

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