BZOJ5100 POI2018Plan metra（构造）

时间：2023-03-09 20:43:32
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dx,1+dx,n的最小值。若为前者，需要满足所有|d1-dn|都相等，挂两棵菊花即可。若为后者，将所有满足dx,1+dx,n=d1,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dx,1-dx,n=dy,1-dy,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 ;

}

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