Codeforces 585.D Lizard Era: Beginning

D. Lizard Era: Beginning

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

In the game Lizard Era: Beginning the protagonist will travel with three companions: Lynn, Meliana and Worrigan. Overall the game has nmandatory quests. To perform each of them, you need to take exactly two companions.

The attitude of each of the companions to the hero is an integer. Initially, the attitude of each of them to the hero of neutral and equal to 0. As the hero completes quests, he makes actions that change the attitude of the companions, whom he took to perform this task, in positive or negative direction.

Tell us what companions the hero needs to choose to make their attitude equal after completing all the quests. If this can be done in several ways, choose the one in which the value of resulting attitude is greatest possible.

Input

The first line contains positive integer n (1 ≤ n ≤ 25) — the number of important tasks.

Next n lines contain the descriptions of the tasks — the i-th line contains three integers li, mi, wi — the values by which the attitude of Lynn, Meliana and Worrigan respectively will change towards the hero if the hero takes them on the i-th task. All the numbers in the input are integers and do not exceed 107 in absolute value.

Output

If there is no solution, print in the first line "Impossible".

Otherwise, print n lines, two characters is each line — in the i-th line print the first letters of the companions' names that hero should take to complete the i-th task ('L' for Lynn, 'M' for Meliana, 'W' for Worrigan). Print the letters in any order, if there are multiple solutions, print any of them.

Examples

input

3
1 0 0
0 1 0
0 0 1

output

LM
MW
MW

input

7
0 8 9
5 9 -2
6 -8 -7
9 4 5
-4 -9 9
-4 5 2
-6 8 -7

output

LM
MW
LM
LW
MW
LM
LW

input

2
1 0 0
1 1 0

output

Impossible

大致题意：一开始3个人的数值为0，有n次操作，每次选两个人将对应的数值增加，最后要求3个人的数值相等.输出方案，如果有多种方案，输出所有数值最大的一种.

分析：n非常小，但是直接搜的话3^25还是会炸.观察到如果指数减小一半，就刚好在能接受的复杂度里了.很容易想到meet in the middle.

应用meet in the middle的方程模型，先列出方程：a + d = b + e = c + f,也就是a - b = e - d 且 a - c = f - d.那么从开头搜，搜到中点就将a-b和a-c的结果放到map里，接着从末尾搜，如果找到了e-d和f-d的值，就更新答案.

两次dfs用三进制数记录答案，注意：第二次dfs记录的答案要倒着输出!

#include <map>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int inf = 0x7fffffff;

int n,A[],B[],C[],maxn,ans = -inf,printx,printy,num[],cntt,cnt;
struct node
{
    int zhuangtai,x;
}e[];

struct node2
{
    int x,y;
    bool operator < (const node2 &a) const
    {
        return a.x == x ? a.y < y : a.x < x;
    }
};

map <node2,int> a;
void dfs1(int dep,int x,int y,int z,int sta)
{
    if (dep == maxn + )
    {
        y -= x;
        z -= x;
        node2 temp;
        temp.x = y;
        temp.y = z;
        if (a.find(temp) == a.end())
        {
            a[temp] = ++cnt;
            e[cnt].zhuangtai = sta;
            e[cnt].x = x;
        }
        else
        {
            if (e[a[temp]].x < x)
            {
                e[a[temp]].x = x;
                e[a[temp]].zhuangtai = sta;
            }
        }
        return;
    }
    for (int i = ; i < ; i++)
    {
        if (i == )
            dfs1(dep + ,x + A[dep],y + B[dep],z,sta * );
        if (i == )
            dfs1(dep + ,x + A[dep],y,z + C[dep],sta *  + );
        if (i == )
            dfs1(dep + ,x,y + B[dep],z + C[dep],sta *  + );
    }
}

void dfs2(int dep,int x,int y,int z,int sta)
{
    if (dep == maxn)
    {
        y -= x;
        z -= x;
        y = -y;
        z = -z;
        node2 temp;
        temp.x = y;
        temp.y = z;
        if (a.find(temp) != a.end())
        {
            if (e[a[temp]].x + x > ans)
            {
                ans = e[a[temp]].x + x;
                printx = e[a[temp]].zhuangtai;
                printy = sta;
            }
        }
        return;
    }
    for (int i = ; i < ; i++)
    {
        if (i == )
            dfs2(dep - ,x + A[dep],y + B[dep],z,sta * );
        if (i == )
            dfs2(dep - ,x + A[dep],y,z + C[dep],sta *  + );
        if (i == )
            dfs2(dep - ,x,y + B[dep],z + C[dep],sta *  + );
    }
}

void print()
{
    for (int i = ; i <= maxn; i++)
    {
        num[++cntt] = printx % ;
        printx /= ;
    }
    for (int i = cntt; i >= ; i--)
    {
        if (num[i] == )
            puts("LM");
        if (num[i] == )
            puts("LW");
        if (num[i] == )
            puts("MW");
    }
    cntt = ;
    for (int i = maxn + ; i <= n; i++)
    {
        num[++cntt] = printy % ;
        printy /= ;
    }
    for (int i = ; i <= cntt; i++)
    {
        if (num[i] == )
            puts("LM");
        if (num[i] == )
            puts("LW");
        if (num[i] == )
            puts("MW");
    }
}

int main()
{
    scanf("%d",&n);
    for (int i = ; i <= n; i++)
        scanf("%d%d%d",&A[i],&B[i],&C[i]);
    if (n %  == )
        maxn = n / ;
    else
        maxn = n /  + ;
    dfs1(,,,,);
    dfs2(n,,,,);
    if (ans != -inf)
    print();
    else
        puts("Impossible");

    return ;
}