cf786a

---

title: CodeForces 786A Berzerk

data: 2018-3-3 10:29:40

tags:

• 博弈论
• bfs
• 无限
• with draw
  
  copyright: true
  
  categories:
• 信息学竞赛
• 题目
  
  description: 有一个环, 1的位置是黑洞, 有一个怪物在任意位置上, 两个人依次行动, 可以将怪物移动自己决策集合中的任意步, 谁先把它送进黑洞谁就赢, 请输出当怪物在任意位置和谁先手的各自的胜负情况, 有平局.

---

A. Berzerk

description

　　Rick and Morty are playing their own version of Berzerk (which has nothing in common with the famous Berzerk game). This game needs a huge space, so they play it with a computer.

　　In this game there are n objects numbered from 1 to n arranged in a circle (in clockwise order). Object number 1 is a black hole and the others are planets. There's a monster in one of the planet. Rick and Morty don't know on which one yet, only that he's not initially in the black hole, but Unity will inform them before the game starts. But for now, they want to be prepared for every possible scenario.

　　Each one of them has a set of numbers between 1 and n - 1 (inclusive). Rick's set is s1 with k1 elements and Morty's is s2 with k2 elements. One of them goes first and the player changes alternatively. In each player's turn, he should choose an arbitrary number like x from his set and the monster will move to his x-th next object from its current position (clockwise). If after his move the monster gets to the black hole he wins.

　　Your task is that for each of monster's initial positions and who plays first determine if the starter wins, loses, or the game will stuck in an infinite loop. In case when player can lose or make game infinity, it more profitable to choose infinity game.

Input

　　The first line of input contains a single integer n (\(2 ≤ n ≤ 7000\)) — number of objects in game.

　　The second line contains integer k1 followed by k1 distinct integers s1, 1, s1, 2, ..., s1, k1 — Rick's set.

　　The third line contains integer k2 followed by k2 distinct integers s2, 1, s2, 2, ..., s2, k2 — Morty's set

\(1 ≤ ki ≤ n - 1\ and\ 1 ≤ si, 1, si, 2, ..., si, ki ≤ n - 1\ for\ 1 ≤ i ≤ 2.\)

Output

　　In the first line print n - 1 words separated by spaces where i-th word is "Win" (without quotations) if in the scenario that Rick plays first and monster is initially in object number i + 1 he wins, "Lose" if he loses and "Loop" if the game will never end.

　　Similarly, in the second line print n - 1 words separated by spaces where i-th word is "Win" (without quotations) if in the scenario that Morty plays first and monster is initially in object number i + 1 he wins, "Lose" if he loses and "Loop" if the game will never end.

Examples

input #1

5

2 3 2

3 1 2 3

output #1

Lose Win Win Loop

Loop Win Win Win

input #2

8

4 6 2 3 4

2 3 6

output #2

Win Win Win Win Win Win Win

Lose Win Lose Lose Win Lose Lose

Way:

　　这显然是一个无限博弈问题, 因为两人都选择最佳策略, 所以两个人有可能一直僵持下去谁也奈何不了谁, 这显然就是平局.

　　显然可以用搜索搜出每个状态的所有后继状态, 根据必败态和必胜态的定义.

• 终止状态是必败态.
• 有一个后继状态是必败态的状态是必胜态.
• 所以后继状态都是必胜态的状态是必败态.

　　通过这样可以推出所有的必胜态和必败态.可是平局如何定义呢?后继状态有不确定状态的状态是不确定状态, 可以发现它构成了一个环.

Code

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<queue>
#define N 7005
using namespace std;

namespace Input{
    inline int read(){
        char ch=getchar();int s=0;
        for(;!isdigit(ch);ch=getchar());
        for(;isdigit(ch);s=s*10+ch-'0',ch=getchar());
        return s;
    }
    inline void read(int &s){
        char ch=getchar();
        for(;!isdigit(ch);ch=getchar());
        for(s=0;isdigit(ch);s=s*10+ch-'0',ch=getchar());
    }
};using namespace Input;
struct Move{
    int pos;
    bool now;
    Move(int _pos=0,bool _now=1){pos=_pos;now=_now;}
};

int mark[N][2],alt[N][2];
int vis[N][2],du[N][2];
int f[N][2],s[2],n;

int mov(int pos,int step){
    return (pos-step+n)%n;
}
void bfs(){
    queue<Move>que;
    f[0][1]=f[0][0]=2;
    que.push((Move){0,0});
    que.push((Move){0,1});
    vis[0][1]=true;
    vis[0][0]=true;
    while(!que.empty()){
        Move top=que.front();que.pop();
        bool opt=top.now^1;
        for(int i=0;i<s[opt];++i){
            int to=mov(top.pos,alt[i][opt]);
            if(f[top.pos][top.now]==1){
                mark[to][opt]++;
                if(!vis[to][opt]&&mark[to][opt]==s[opt]){
                    f[to][opt]=2;
                    vis[to][opt]=true;
                    que.push((Move){to,opt});
                }
            }
            else if(!vis[to][opt]){
                f[to][opt]=1;
                vis[to][opt]=true;
                que.push((Move){to,opt});
            }
        }
    }
}

int main(){
    read(n);
    read(s[0]);
    for(int i=0;i<s[0];++i)read(alt[i][0]);
    read(s[1]);
    for(int i=0;i<s[1];++i)read(alt[i][1]);
    bfs();int ans;
    for(int i=1;i<n;++i){
        ans=f[i][0];
        printf(ans?ans==1?"Win":"Lose":"Loop");
        putchar(' ');
    }
    putchar('\n');
    for(int i=1;i<n;++i){
        ans=f[i][1];
        printf(ans?ans==1?"Win":"Lose":"Loop");
        putchar(' ');
    }
    system("pause");
    return 0;
}