BZOJ5100 POI2018Plan metra（构造）

　　容易发现要么1和n直接相连，要么两点距离即为所有d_x,1+d_x,n的最小值。若为前者，需要满足所有|d₁-d_n|都相等，挂两棵菊花即可。若为后者，将所有满足d_x,1+d_x,n=d_1,n的挂成一条链，其余点直接与链上点相连即可，相连点需要满足d_x,1-d_x,n=d_y,1-d_y,n。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 500010
#define D 1000010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int n,a[N],b[N],id[D];
struct data
{
    int x,i;
    bool operator <(const data&a) const
    {
        return x<a.x;
    }
}d[N];
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj5100.in","r",stdin);
    freopen("bzoj5100.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read();
    for (int i=;i<n;i++) a[i]=read();
    for (int i=;i<n;i++) b[i]=read();
    if (n==) {cout<<"TAK\n"<<endl<<<<' '<<<<' '<<;return ;}
    int len=abs(a[]-b[]);
    for (int i=;i<n;i++) if (abs(a[i]-b[i])!=len) {len=;break;}
    if (len)
    {
        cout<<"TAK\n";
        cout<<<<' '<<n<<' '<<len<<endl;
        for (int i=;i<n;i++)
        if (a[i]>b[i]) printf("%d %d %d\n",i,n,b[i]);
        else printf("%d %d %d\n",i,,a[i]);
        return ;
    }
    len=;
    for (int i=;i<n;i++) len=min(len,a[i]+b[i]);
    int cnt=;for (int i=;i<n;i++) if (a[i]+b[i]==len) cnt++,d[cnt].x=a[i],d[cnt].i=i;
    sort(d+,d+cnt+);++cnt;d[].x=,d[].i=,d[cnt].x=len,d[cnt].i=n;
    for (int i=;i<=cnt;i++) if (d[i].x==d[i-].x) {cout<<"NIE";return ;}
    for (int i=;i<=cnt;i++) id[d[i].x]=d[i].i;
    for (int i=;i<n;i++)
    if (a[i]+b[i]>len&&((a[i]+b[i]-len&)||!id[a[i]-(a[i]+b[i]-len>>)])) {cout<<"NIE";return ;}
    cout<<"TAK\n";
    for (int i=;i<=cnt;i++) printf("%d %d %d\n",d[i-].i,d[i].i,d[i].x-d[i-].x);
    for (int i=;i<n;i++)
    if (a[i]+b[i]>len) printf("%d %d %d\n",id[a[i]-(a[i]+b[i]-len>>)],i,a[i]+b[i]-len>>);
    return ;
}