「日常训练」「小专题·USACO」 Wormholes（1-4）

题意

之后补充。

分析

这是一条很好的考察递归（或者说搜索）的题目。它的两个过程（建立初步解，验证）都用到了递归（或者说运用递归可以相当程度的减少代码量）。

具体实现见代码。注意，为了使用std::pair的比较操作符，代码交换了x、y的位置。

代码

/*
ID: samhx1
LANG: C++14
TASK: wormhole
*/
#include <bits/stdc++.h>
#define MP make_pair
#define PB push_back
#define fi first
#define se second
#define ZERO(x) memset((x), 0, sizeof(x))
#define ALL(x) (x).begin(),(x).end()
#define rep(i, a, b) for (int i = (a); i <= (b); ++i)
#define per(i, a, b) for (int i = (a); i >= (b); --i)
#define QUICKIO                  \
    ios::sync_with_stdio(false); \
    cin.tie(0);                  \
    cout.tie(0);
using namespace std;
using ll = long long;
using ull = unsigned long long;
using pi = pair<ll,ll>;
using pii = pair<ll, pi>;
// Need to write editorial about it.
template<typename T>
T readint()
{
    T tmp; cin>>tmp;
    return tmp;
}
int n,ans=0;
pi pnt[15];
ll nxt[15],partner[15];
int vis[15];
bool check(int spnt)
{
    if(vis[spnt]>=2) return true;
    vis[spnt]++;
    if(nxt[spnt]==-1) return false;
    else
    {
        vis[nxt[spnt]]++;
        rep(i,1,n)
        {
            if(i!=nxt[spnt] && partner[nxt[spnt]]==partner[i])
            {
                return check(i);
                break; // should not be executed.
            }
        }
    }
    return false;
}
bool check()
{
    for(int nowpnt=1;nowpnt<=n;++nowpnt)
    {
        memset(vis,0,sizeof(vis));
        if(check(nowpnt)) return true;
    }
    return false;
}
void solve(int dep)
{
    if(dep==n/2)
    {
        rep(i,1,n) if(partner[i]==-1) return;
        if(check())
        {
            //rep(i,1,n) cout<<partner[i]<<" ";
            //cout<<endl;
            //cout<<"Success!"<<endl;
            ans++;
        }
    }
    else
    {
        rep(i,1,n)
        {
            if(partner[i]==-1)
            {
                partner[i]=dep+1;
                rep(j,i+1,n)
                {
                    if(partner[j]!=-1) continue;
                    partner[j]=dep+1;
                    //printf("i=%d, j=%d: ",i,j);
                    //rep(k,1,n) cout<<partner[k]<<" ";
                    //cout<<endl;
                    solve(dep+1);
                    partner[j]=-1;
                }
                partner[i]=-1;
                break;
            }
        }
    }
}
int main()
{
    freopen("wormhole.in","r",stdin);
    freopen("wormhole.out","w",stdout);
    cin>>n;
    rep(i,1,n)
    {
        int x,y; cin>>x>>y;
        pnt[i]=MP(y,x); // WARN: x,y has been swapped.
    }
    sort(pnt+1,pnt+n+1);
    memset(nxt,-1,sizeof(nxt));
    rep(i,1,n)
    {
        if(i==n) continue;
        if(pnt[i+1].fi==pnt[i].fi) nxt[i]=i+1;
    }
    memset(partner,-1,sizeof(partner));
    solve(0);
    cout<<ans<<endl;
    return 0;
}