(中等) CF 585D Lizard Era: Beginning，中途相遇。

　　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.

 

　　题目就是问能不能找到一组序列然后符合条件。。。

　　一眼看去就是中途相遇，n只有25，两个dfs就行了。。。

　　第一次用SBT维护的，然后1996ms险过，第二次改成sort，498ms，不得不说真是快啊。。。

 

代码如下：

// ━━━━━━神兽出没━━━━━━
//      ┏┓       ┏┓
//     ┏┛┻━━━━━━━┛┻┓
//     ┃           ┃
//     ┃     ━     ┃
//     ████━████   ┃
//     ┃           ┃
//     ┃    ┻      ┃
//     ┃           ┃
//     ┗━┓       ┏━┛
//       ┃       ┃
//       ┃       ┃
//       ┃       ┗━━━┓
//       ┃           ┣┓
//       ┃           ┏┛
//       ┗┓┓┏━━━━━┳┓┏┛
//        ┃┫┫     ┃┫┫
//        ┗┻┛     ┗┻┛
//
// ━━━━━━感觉萌萌哒━━━━━━

// Author        : WhyWhy
// Created Time  : 2015年10月13日 星期二 17时27分11秒
// File Name     : F.cpp

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>

using namespace std;

const int MaxN=2000005;
const int INF=0x3f3f3f3f;

struct Num
{
    int a,b;

    Num() {}
    Num(int x,int y):a(x),b(y) {}

    bool operator < (const Num & y) const
    {
        return a==y.a ? b<y.b : a<y.a;
    }

    bool operator > (const Num & y) const
    {
        return a==y.a ? b>y.b : a>y.a;
    }

    bool operator == (const Num & y) const
    {
        return a==y.a && b==y.b;
    }

    Num operator + (const Num & y) const
    {
        return Num(a+y.a,b+y.b);
    }
};

struct State
{
    Num n;
    int base;
    long long route;

    bool operator < (const State &b) const
    {
        return n<b.n;
    }

    bool operator == (const State &b) const
    {
        return n==b.n;
    }
};

bool cmp(const State &a,const State &b)
{
    return a.n==b.n ? a.base>b.base : a.n<b.n;
}

State sta[MaxN];
int cou;

int N;
Num rem[30][3];
int rnum[30];

int bbb;
long long lll;

void dfs1(int d,Num tn,int b,long long l)
{
    if(d>((N+1)>>1))
    {
        sta[cou].n=tn;
        sta[cou].base=b;
        sta[cou++].route=l;
        return;
    }

    for(int i=0;i<3;++i)
        dfs1(d+1,tn+rem[d][i],b+(i ? rnum[d] : 0),(l<<2)|i);
}

void dfs2(int d,Num tn,int b,long long l)
{
    if(d>N)
    {
        State ts;
        ts.n.a=-tn.a;
        ts.n.b=-tn.b;

        int p=lower_bound(sta,sta+cou,ts)-sta;

        if(p<cou && sta[p]==ts && sta[p].base+b>bbb)
        {
            bbb=sta[p].base+b;
            lll=(sta[p].route<<(2*(N/2)))|l;
        }

        return;
    }

    for(int i=0;i<3;++i)
        dfs2(d+1,tn+rem[d][i],b+(i ? rnum[d] : 0),(l<<2)|i);
}

void show()
{
    int rrr[30];

    for(int i=0;i<N;++i)
    {
        rrr[i]=lll & 3;
        lll>>=2;
    }

    for(int i=N-1;i>=0;--i)
        if(rrr[i]==0) puts("MW");
        else if(rrr[i]==1) puts("LW");
        else puts("LM");
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

    int a,b,c;

    scanf("%d",&N);
    for(int i=1;i<=N;++i)
    {
        scanf("%d %d %d",&a,&b,&c);
        rnum[i]=a;
        rem[i][0]=Num(b,c);
        rem[i][1]=Num(-a,c-a);
        rem[i][2]=Num(b-a,-a);
    }

    cou=0;
    bbb=-INF;
    dfs1(1,Num(0,0),0,0);
    sort(sta,sta+cou,cmp);
    cou=unique(sta,sta+cou)-sta;
    dfs2((N+1)/2+1,Num(0,0),0,0);

    if(bbb==-INF) puts("Impossible");
    else show();
    
    return 0;
}

View Code