Description
Statements
In a galaxy far, far away the great galactic empire had collapsed after the death of the emperor, and now a new corrupted galactic republic gradually grows stronger. But the evil Sith don't spend time in vain: they've created opposition named «The last order» and now they are planning to choose a leader by a fair tournament.
In the tournament there will be n Siths who has left alive. The i-th of these Siths has three integer characteristics: mastery of lightsabers ai, mastery of the Force bi and the level of mental balance ci. All characteristics of all Siths are different. The tournament consists of one on one fights. The winner of each of these fights is the Sith who has at least two characteristics that are bigger than their opponent's corresponding ones.
The problem is that the Sith tournaments have never been organized to find the strongest Sith who could beat anyone else: instead, they were supposed to distinguish the weakest one, somebody who couldn't win any fight, and to let everybody take a good laugh at that loser. It seems like such Sith will become the leader of «The last order», so it'd be good to know them in advance.
Input
The first line contains a single integer n (2 ≤ n ≤ 2·105) — the number of the Siths who will participate in the tournament.
Each of the next n lines contains three space-separated integers: ai, bi, ci (1 ≤ ai, bi, ci ≤ 109) — the three corresponding characteristics of the i-th Sith.
Output
In the first line output a single integer — the number of Siths who can't win any fight.
In the second line print the numbers of those Siths separated by a space.
Sample Input
4 3 11 9 1 4 8 5 2 10 12 7 6
1 2
3 700 40 1 50 2 800 3 900 60
0
排序，因为必有输赢，所以答案只有1和0两种情况
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=5e2+10;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;
while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')
fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,
rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
typedef struct node{
ll id;
ll a;
ll b;
ll c;
}NODE;
NODE a[1000005];
bool cmp(NODE x,NODE y)
{
int ct=0;
if(x.a>y.a)
ct ++;
if(x.b>y.b)
ct ++;
if(x.c>y.c)
ct ++;
if(ct>=2)
return true;
else
return false;
}
int main()
{
ll n;
cin >> n;
for(int i=0;i<n;i++)
{
cin >> a[i].a >> a[i].b >> a[i].c ;
a[i].id = i+1;
}
sort(a,a+n,cmp);
NODE t=a[n-1];
int i;
for(i=0;i<n-1;i++)
{
if(cmp(t,a[i]))
{
//cout << a[i].id << endl;
break;
}
}
if(i==n-1)
{
printf("1\n");
printf("%lld\n",a[n-1].id);
}
else
{
printf("0\n");
}
return 0;
}