Codeforces Round #325 (Div. 2) F. Lizard Era: Beginning meet in the mid

F. Lizard Era: Beginning

Time Limit: 1 Sec  

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/586/problem/F

Description

In the game Lizard Era: Beginning the protagonist will travel with three companions: Lynn, Meliana and Worrigan. Overall the game has n mandatory 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 l_i, m_i, w_i — 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.

​​,C​i​​，即此题的初始分值、每分钟减少的分值、dxy做这道题需要花费的时间。

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.

Sample Input

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

Sample Output

LM
MW
LM
LW
MW
LM
LW

HINT

题意

有n个任务，有三个人

每次任务都必须派两个人出去，每个任务都会使得人涨能力值，不同人涨的不一样

然后问你有没有一种方案可以使得所有人最后的能力值都一样

如果有多种方案，请输出可以最后使得能力值最大的一种方案

题解：

meet in the mid,状态存三个量就好了A，A-B，B-C，然后就可以瞎搜了，注意最后要输出方案，所以就直接把中间的过程都状压一下就行了

代码：

#include<iostream>
#include<stdio.h>
#include<map>
#include<vector>
using namespace std;

int mid,n;
int a[30],b[30],c[30];
map<pair<int,int> ,pair<int,long long> >H;
pair<int,long long> ans;
void dfs1(int step,long long sta,int A,int B,int C)
{
    if(step==mid+1)
    {
        pair<int,long long> temp;
        temp = H[pair<int,int>(A-B,B-C)];
        if(temp.second == 0)
            H[pair<int,int>(A-B,B-C)] = pair<int,long long>(A,sta);
        else
            H[pair<int,int>(A-B,B-C)] = max(temp,pair<int,long long>(A,sta));
        return;
    }
    dfs1(step+1,sta<<2|1,A,B+b[step],C+c[step]);
    dfs1(step+1,sta<<2|2,A+a[step],B,C+c[step]);
    dfs1(step+1,sta<<2|3,A+a[step],B+b[step],C);
}
void dfs2(int step,long long sta,int A,int B,int C)
{
    if(step==n+1)
    {
        pair<int,long long> temp = H[pair<int,int>(B-A,C-B)];
        if(temp.second==0)return;
        ans = max(ans,pair<int,long long>(A+temp.first,temp.second<<(n-mid<<1)|sta));
        return;
    }
    dfs2(step+1,sta<<2|1,A,B+b[step],C+c[step]);
    dfs2(step+1,sta<<2|2,A+a[step],B,C+c[step]);
    dfs2(step+1,sta<<2|3,A+a[step],B+b[step],C);
}
int main()
{
    ans.first = -99999999;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d%d%d",&a[i],&b[i],&c[i]);
    mid = (n+1)/2;
    dfs1(1,0,0,0,0);
    dfs2(mid+1,0,0,0,0);
    if(ans.first == -99999999)return puts("Impossible");
    vector<int> Ans;
    for(int i=1;i<=n;i++)
    {
        Ans.push_back(ans.second&3);
        ans.second>>=2;
    }
    for(int i=Ans.size()-1;i>=0;i--)
    {
        if(Ans[i]==3)
            cout<<"LM"<<endl;
        else if(Ans[i]==2)
            cout<<"LW"<<endl;
        else
            cout<<"MW"<<endl;
    }

}