题目链接在这里：2017-2018 ACM-ICPC, Asia Tsukuba Regional Contest
这道题首先学习的是平行线的判断，不要直接用double存斜率，应该存分子分母，然后除以gcd化成最简。
然后学习了一下dfs爆搜匹配的方法，固定了一个点以后直接往后找就行了，不要每次都从头再找
1 #include "bits/stdc++.h" 2 using namespace std; 3 typedef long long LL; 4 const LL MAX=25; 5 LL n,ans,x[MAX],y[MAX],used[MAX],cnt; 6 bool vis[MAX]; 7 LL gcd(LL xx,LL yy){return yy==0?xx:gcd(yy,xx%yy);} 8 pair <LL,LL> k[MAX]; 9 bool cmp(pair<LL,LL> xx,pair<LL,LL> yy){ 10 if (xx.first!=yy.first) return xx.first<yy.first; 11 return xx.second<yy.second; 12 } 13 void check(){ 14 LL i,j,zt,zt2,gg; 15 zt2=0,cnt=0; 16 pair <LL,LL> kk; 17 memset(vis,false,sizeof(vis)); 18 for (i=1;i<=n;i++){ 19 if (vis[i]) continue; 20 vis[i]=vis[used[i]]=true; 21 kk.first=y[i]-y[used[i]]; 22 kk.second=x[i]-x[used[i]]; 23 gg=gcd(abs(kk.first),abs(kk.second)); 24 if (gg!=0) kk.first/=gg,kk.second/=gg; 25 if (kk.first<0) kk.first*=-1,kk.second*=-1; 26 if (kk.first==0) kk.second=abs(kk.second); 27 if (kk.second==0) kk.first=abs(kk.first); 28 k[++cnt]=kk; 29 } 30 //for (i=1;i<=n;i++) cout<<used[i]<<' ';cout<<endl; 31 sort(k+1,k+cnt+1,cmp); 32 //for (i=1;i<=cnt;i++) cout<<k[i].first<<' '<<k[i].second<<endl; 33 //cout<<endl; 34 zt=1; 35 for (i=2;i<=cnt+1;i++){ 36 if (k[i]!=k[i-1]){ 37 zt2+=zt*(zt-1)/2; 38 zt=1; 39 } 40 else zt++; 41 } 42 ans=max(ans,zt2); 43 } 44 void dfs(int a,int b){ 45 if(a==n) return; 46 if(used[a]) dfs(a+1,b); 47 for(int i=a+1;i<=n;i++){ 48 if((!used[i])&&(!used[a])){ 49 used[a]=i; 50 used[i]=a; 51 if(b==n/2) check(); 52 else dfs(a+1,b+1); 53 used[a]=0; 54 used[i]=0; 55 } 56 } 57 } 58 int main(){ 59 freopen ("b.in","r",stdin); 60 freopen ("b.out","w",stdout); 61 LL i,j,gg; 62 scanf("%lld",&n); 63 for (i=1;i<=n;i++) 64 scanf("%lld%lld",x+i,y+i); 65 memset(vis,false,sizeof(vis)); 66 dfs(1,1); 67 printf("%lld",ans); 68 return 0; 69 }