Codeforces Round #378 (Div. 2) -- D. Kostya the Sculptor （STL水题）

大体题意：

你有n 个长方体形状的石头，你要送给你的朋友，你做多选择两个石头，要求石头内部的圆的半径尽可能大！输出选择石头的标号？选择两个石头的条件是 两个石头必须能连接起来！

思路：

很水的一道题目，却因为C题 没有做这个题目！

我们直接开一个map<pair<int,int> ,multiset<Node> >mp;

第一个pair 记录 你选择长方体的面，  第二个记录 高，  先枚举只选一个石头的情况，在枚举选择两个石头的情况！

取个最大值即可!

之所以  第二个要开 multiset 是因为题目中所说  有可能两个石头的长宽高都一样，但这依然算两种石头！

可以简单的估算一下复杂度，最多10w 个长方体，那么第一个pair 数量最多也就是30W个！

所以说枚举不可能会超时的！

详细见代码：

#include <bits/stdc++.h>#define mr make_pair#define fi first#define se second#define Min(a,b) (a) > (b) ? (b) : (a)#define Max(a,b) (a) > (b) ? (a) : (b)using namespace std;struct Node{    int v;    int id;    Node(int v = 0,int id = 0):v(v),id(id){}    bool operator < (const Node& rhs) const {        return v > rhs.v;    }};map<pair<int,int> ,multiset<Node> >mp;map<pair<int,int> ,multiset<Node> >::iterator it;multiset<Node>::iterator it2;void add(int a,int b,int c,int id){    if (b > c) swap(b,c);    mp[mr(b,c)].insert(Node(a,id));}int main(){    int n;    scanf("%d",&n);    int Maxr = 0;    pair<int,int>pos;    pos.se = -1;    for (int i = 0; i < n; ++i){        int a,b,c;        scanf("%d %d %d",&a, &b, &c);        int m = Min(a,Min(b,c));        if (m > Maxr){            Maxr = m;            pos.fi = i+1;        }        add(a,b,c,i+1);        add(b,a,c,i+1);        add(c,b,a,i+1);    }    for (it = mp.begin(); it != mp.end(); ++it){        pair<int,int>p = it->fi;        multiset<Node>st = it->se;        if ((int)st.size() < 2) continue;        int sum = 0,cnt = 0;        int tmp[3];        for (it2 = st.begin(),cnt = 0; cnt<2 && it2 != st.end(); ++cnt,++it2){            sum += (*it2).v;            tmp[cnt] = (*it2).id;        }        int m = Min(sum,Min(p.fi,p.se));        if (m > Maxr){            m = Maxr;            pos.fi = tmp[0];            pos.se = tmp[1];        }    }    if (~pos.se ) printf("2\n%d %d\n",pos.fi,pos.se);    else printf("1\n%d\n",pos.fi);    return 0;}

D. Kostya the Sculptor

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

input

65 5 53 2 41 4 12 1 33 2 43 3 4

output

11

input

710 7 85 10 34 2 65 5 510 2 84 2 17 7 7

output

21 5

Note

In the first example we can connect the pairs of stones:

• 2 and 4, the size of the parallelepiped: 3 × 2 × 5, the radius of the inscribed sphere 1
• 2 and 5, the size of the parallelepiped: 3 × 2 × 8 or 6 × 2 × 4 or 3 × 4 × 4, the radius of the inscribed sphere 1, or 1, or 1.5respectively.
• 2 and 6, the size of the parallelepiped: 3 × 5 × 4, the radius of the inscribed sphere 1.5
• 4 and 5, the size of the parallelepiped: 3 × 2 × 5, the radius of the inscribed sphere 1
• 5 and 6, the size of the parallelepiped: 3 × 4 × 5, the radius of the inscribed sphere 1.5

Or take only one stone:

• 1 the size of the parallelepiped: 5 × 5 × 5, the radius of the inscribed sphere 2.5
• 2 the size of the parallelepiped: 3 × 2 × 4, the radius of the inscribed sphere 1
• 3 the size of the parallelepiped: 1 × 4 × 1, the radius of the inscribed sphere 0.5
• 4 the size of the parallelepiped: 2 × 1 × 3, the radius of the inscribed sphere 0.5
• 5 the size of the parallelepiped: 3 × 2 × 4, the radius of the inscribed sphere 1
• 6 the size of the parallelepiped: 3 × 3 × 4, the radius of the inscribed sphere 1.5

It is most profitable to take only the first stone.