Minimum Distance in a Star Graph BFS/康托展开 2017 ACM-ICPC 亚洲区（南宁赛区）网络赛

题目链接

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模拟+BFS+康拓展开,此题的数据比较水,如果n为9就GG了

#include<bits/stdc++.h>using namespace std;typedef long long ll;vector<int> mm[500005];int vis[500005];int n;ll sum =0;int a[11];int b[11];int  fac[] = {1,1,2,6,24,120,720,5040,40320}; //i的阶乘为fac[i]// 康托展开-> 表示数字a是 a的全排列中从小到大排，排第几// n表示1~n个数  a数组表示数字。int kangtuo(){    int i,j,t,sum;    sum=0;    for( i=0; i<n ;++i)    {        t=0;        for(j=i+1;j<n;++j)            if( a[i]>a[j] )                ++t;        sum+=t*fac[n-i-1];    }    return sum+1;}struct NODE{    int x;    int step;};int  bfs(int ss,int tt){    queue <NODE> que;    memset(vis,0,sizeof(vis));    NODE cur,next;    cur.x = ss;    cur.step = 0;    vis[cur.x] = 1;    que.push(cur);    while(!que.empty())    {        cur = que.front();        que.pop();        if(cur.x==tt)        {            return cur.step;        }        for(int i=0;i<mm[cur.x].size();i++)        {            if(vis[ mm[cur.x][i]]) continue;            next = cur;            next.x = mm[cur.x][i];            next.step +=1;            vis[next.x] = 1;            que.push(next);        }    }}int main(){//    freopen("data.txt","r",stdin);    scanf("%d",&n);    for(int i=0;i<n;i++)    {        a[i] = b[i] = i+1;    }    sum = 1;    for(int i=1;i<=n;i++)    {        sum *= i;    }    for(int i=1;i<=sum;i++)    {        for(int j=0;j<n;j++)        {            a[j] = b[j];        }        for(int j=1;j<n;j++)        {            swap(a[0],a[j]);            ll to = kangtuo();            mm[i].push_back(to);            mm[to].push_back(i);            swap(a[0],a[j]);        }        next_permutation(b,b+n);    }    ll s,t;    for(int i=0;i<5;i++)    {        scanf("%lld%lld",&s,&t);        int dig=n-1;        while(s)        {            a[dig--] = s%10;            s/=10;        }        s= kangtuo();        dig=n-1;        while(t)        {            a[dig--] = t%10;            t/=10;        }        t= kangtuo();        printf("%d\n",bfs(s,t));    }    return 0;}