Kostya is a genial sculptor, he has an idea: to carve a marble sculpture in the shape of a sphere. Kostya has a friend Zahar who works at a career. Zahar knows about Kostya‘s idea and wants to present him a rectangular parallelepiped of marble from which he can carve the sphere.
Zahar has n stones which are rectangular parallelepipeds. The edges sizes of the i-th of them are ai, bi and ci. He can take no more than two stones and present them to Kostya.
If Zahar takes two stones, he should glue them together on one of the faces in order to get a new piece of rectangular parallelepiped of marble. Thus, it is possible to glue a pair of stones together if and only if two faces on which they are glued together match as rectangles. In such gluing it is allowed to rotate and flip the stones in any way.
Help Zahar choose such a present so that Kostya can carve a sphere of the maximum possible volume and present it to Zahar.
Input
The first line contains the integer n (1 ≤ n ≤ 105).
n lines follow, in the i-th of which there are three integers ai, bi and ci (1 ≤ ai, bi, ci ≤ 109) — the lengths of edges of the i-th stone. Note, that two stones may have exactly the same sizes, but they still will be considered two different stones.
Output
In the first line print k (1 ≤ k ≤ 2) the number of stones which Zahar has chosen. In the second line print k distinct integers from 1 to n — the numbers of stones which Zahar needs to choose. Consider that stones are numbered from 1 to n in the order as they are given in the input data.
You can print the stones in arbitrary order. If there are several answers print any of them.
题意：给你n个矩形，当它们任意一个面积相同时可以合并，最多合并两个，问你组成最大内切求需要哪几个矩形组合，如果一个矩形就能组成
最大那么就输出1然后它的下表，如果需要两个（最多两个）输出2然后它们的下标。
这题作为d题是挺简单的，思路简单构成的矩形内切圆大小取决于它的最小边，所以如果拿最小边那个面去合成的话最多也不会超过最小边，
所以要拿最大边和次大边组合才有可能得到大的。于是处理一下3个边从大到小排序一下，在将这些点从大到小排序一下。先记录一个矩形时
最大结果的下标是多少， 再考虑两个的时候大边于次大边这个面组合，最大内切圆半径为min(sum , s[i].y) （sum表示两最小边之和，s[i].y
表示次小边）。大致就是这样的思路。
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int M = 1e5 + 10;
typedef long long ll;
struct ss {
ll x , y , z , num;
}s[M];
bool cmp(ss a , ss b) {
if(a.x == b.x)
return a.y > b.y;
return a.x > b.x;
}
int main()
{
int n;
cin >> n;
ll MAX = 0;
int temp = 0;
for(int i = 0 ; i < n ; i++) {
ll a , b , c;
cin >> a >> b >> c;
ll sum = a + b + c;
s[i].x = max(a , max(b , c));
s[i].z = min(a , min(b , c));
s[i].y = (sum - s[i].x - s[i].z);
s[i].num = i;
if(MAX < s[i].z) {
temp = i;
MAX = s[i].z;
}
}
sort(s , s + n , cmp);
ll sum2 = 0;
int l = 0;
int r = l;
int l2 = l;
for(int i = 0 ; i < n - 1 ; i++) {
if(s[i].x == s[i + 1].x && s[i].y == s[i + 1].y) {
l = i;
sum2 = s[i].z + s[i + 1].z;
if(MAX < min(sum2 , s[i].y)) {
MAX = min(sum2 , s[i].y);
r = i + 1;
l2 = l;
}
}
}
if(r - l2 >= 1) {
cout << r - l2 + 1 << endl;
for(int i = l2 ; i <= r ; i++) {
cout << s[i]. num + 1 << ‘ ‘;
}
}
else {
cout << 1 << endl;
cout << temp + 1 << endl;
}
return 0;
}