Codeforces 919F. A Game With Numbers（博弈论）

 

Imagine that Alice is playing a card game with her friend Bob. They both have exactly $\mathrm{88 cards\; and\; there\; is\; an\; integer\; on\; each\; card,\; ranging\; from 00 to 44.\; In\; each\; round,\; Alice\; or\; Bob\; in\; turns\; choose\; two\; cards\; from\; different\; players,\; let\; them\; be aa and bb,\; where aa is\; the\; number\; on\; the\; player\text{'}s\; card,\; and bb is\; the\; number\; on\; the\; opponent\text{'}s\; card.\; It\; is\; necessary\; that a\cdot b\ne 0a\cdot b\ne 0.\; Then\; they\; calculate c=(a+b)mod5c=(a+b)mod5 and\; replace\; the\; number aa with cc.\; The\; player\; who\; ends\; up\; with\; numbers\; on\; all 88 cards\; being 00,\; wins.}$

Now Alice wants to know who wins in some situations. She will give you her cards' numbers, Bob's cards' numbers and the person playing the first round. Your task is to determine who wins if both of them choose the best operation in their rounds.

Input

The first line contains one positive integer $\mathrm{TT (1\le T\le 1000001\le T\le 100000),\; denoting\; the\; number\; of\; situations\; you\; need\; to\; consider.}$

The following lines describe those $\mathrm{TT situations.\; For\; each\; situation:}$

• The first line contains a non-negative integer $\mathrm{ff (0\le f\le 10\le f\le 1),\; where f=0f=0 means\; that\; Alice\; plays\; first\; and f=1f=1 means\; Bob\; plays\; first.}$
• The second line contains $\mathrm{88 non-negative\; integers a1,a2,\dots ,a8a1,a2,\dots ,a8 (0\le ai\le 40\le ai\le 4),\; describing\; Alice\text{'}s\; cards.}$
• The third line contains $\mathrm{88 non-negative\; integers b1,b2,\dots ,b8b1,b2,\dots ,b8 (0\le bi\le 40\le bi\le 4),\; describing\; Bob\text{'}s\; cards.}$

We guarantee that if $\mathrm{f=0f=0,\; we\; have \sum 8i=1ai\ne 0\sum i=18ai\ne 0.\; Also\; when f=1f=1, \sum 8i=1bi\ne 0\sum i=18bi\ne 0 holds.}$

Output

Output $\mathrm{TT lines.\; For\; each\; situation,\; determine\; who\; wins.\; Output}$

• "Alice" (without quotes) if Alice wins.
• "Bob" (without quotes) if Bob wins.
• "Deal" (without quotes) if it gets into a deal, i.e. no one wins.

解题思路:

博弈论，假如说做出一个决定，之后做出的可能的决定存在先手必败，那么这个先手一定像那个状态选择，这样后手作为新的新手就一定必败。

而如果后继状态中只要有先手必胜，那么这个人一定尽量不选择这个状态。

将状态抽象成点，将可以转移到的状态之间连上有向边，就出现了一个图。

比如说这道题，可以将可能的状态（495²个）连边，当然我们要反向处理。

确定先手必胜时BFS，否则拓扑排序。

细节要好好处理，二人是不会使用0去更新的（见题目描述）。

代码：

#include<map>
#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
struct pnt{
    std::pair<int,int>sit;
    int hd;
    int ind;
    int fin;//-1 先手必败 
}p[1000000];
struct ent{
    int twd;
    int lst;
}e[10000000];
std::queue<int>Q;
std::map<int,int>M1;
std::map<std::pair<int,int>,int>M2;
int S[10000][10];
int H[10000];
int tmp[10];
int cnt;
int n,m;
int T;
int trans(int *a);
int indx(int sd);
void dfs(int x);
void zip(void);
void build(void);
void Bfs(void);
void markimp(void);
void addedge(void);
void work(void);
void Pre(void);
void ade(int f,int t);
int main()
{
    Pre();
    scanf("%d",&T);
    while(T--)
        work();
    return 0;
}
void Pre(void)
{
    dfs(1);
    zip();
    build();
    Bfs();
    return ;
}
void dfs(int x)
{
    if(x==9)
    {
        m++;
        for(int i=1;i<=8;i++)
        {
            S[m][i]=tmp[i];
            H[m]=H[m]*5+tmp[i];
        }
        M1[H[m]]=m;
        return ;
    }
    for(int i=tmp[x-1];i<=4;i++)
    {
        tmp[x]=i;
        dfs(x+1);
    }
    return ;
}
void zip(void)
{
    for(int i=1;i<=m;i++)
    {
        for(int j=1;j<=m;j++)
        {
            p[++n].sit=std::make_pair(i,j);
            M2[std::make_pair(i,j)]=n;
        }
    }
    return ;
}
void build(void)
{
    markimp();
    addedge();
    return ;
}
void Bfs(void)
{
    while(!Q.empty())
    {
        int x=Q.front();
        Q.pop();
        for(int i=p[x].hd;i;i=e[i].lst)
        {
            int to=e[i].twd;
            if(p[to].ind==0)
                continue;
            if(p[x].fin==-1)
            {
                p[to].ind=0;
                p[to].fin=1;
                Q.push(to);
            }else{
                p[to].ind--;
                if(!p[to].ind&&!p[to].fin)
                {
                    p[to].fin=-1;
                    Q.push(to);
                }
            }
        }
    }
    return ;
}
void markimp(void)
{
    int sta=M1[0];
    for(int i=1;i<=n;i++)
    {
        if(p[i].sit.first==sta)
        {
            p[i].fin=1;
            Q.push(i);
        }else if(p[i].sit.second==sta)
        {
            p[i].fin=-1;
            Q.push(i);
        }
    }
    return ;
}
void addedge(void)
{
    for(int x=1;x<=n;x++)
    {
        if(p[x].fin)
            continue;
        for(int i=1;i<=8;i++)
            tmp[i]=S[p[x].sit.first][i];
        for(int f=1;f<=8;f++)
        {
            if(f!=1&&tmp[f]==tmp[f-1])
                continue;
            if(!tmp[f])
                continue;
            int a=tmp[f];
            for(int t=1;t<=8;t++)
            {
                if(t!=1&&S[p[x].sit.second][t]==S[p[x].sit.second][t-1])
                    continue;
                int b=S[p[x].sit.second][t];
                if(!b)
                    continue;
                int c=(a+b)%5;
                tmp[f]=c;
                int t0=M1[trans(tmp)];
                int y=M2[std::make_pair(p[x].sit.second,t0)];
                ade(y,x);
                for(int i=1;i<=8;i++)
                    tmp[i]=S[p[x].sit.first][i];
            }
        }
    }
}
int trans(int *a)
{
    int ans=0;
    std::sort(a+1,a+9);
    for(int i=1;i<=8;i++)
        ans=ans*5+a[i];
    return ans;
}
void ade(int f,int t)
{
    cnt++;
    e[cnt].twd=t;
    e[cnt].lst=p[f].hd;
    p[f].hd=cnt;
    p[t].ind++;
    return ;
}
void work(void)
{
    int f;
    scanf("%d",&f);
    int t0,t1;
    for(int i=1;i<=8;i++)
        scanf("%d",&tmp[i]);
    t0=M1[trans(tmp)];
    for(int i=1;i<=8;i++)
        scanf("%d",&tmp[i]);
    t1=M1[trans(tmp)];
    if(f)
        std::swap(t0,t1);
    if(f)
    {
        int x=M2[std::make_pair(t0,t1)];
        if(p[x].fin==1)
        {
            puts("Bob");
        }else if(p[x].fin==0)
        {
            puts("Deal");
        }else{
            puts("Alice");
        }
    }else{
        int x=M2[std::make_pair(t0,t1)];
        if(p[x].fin==-1)
        {
            puts("Bob");
        }else if(p[x].fin==0)
        {
            puts("Deal");
        }else{
            puts("Alice");
        }
    }
}