题解 P7336 【[JRKSJ R1] 1+1】

· · 题解

正常的赛时题解大家可以在这里看 https://www.luogu.com.cn/blog/2-6-5-3-5/ti-xie-t411-post

其实还有一个小优化就是那个 dfs 是可以记忆化的。

其他的东西 cmll02 都写了

这里说一下我是怎么优化的

首先大家的做法基本都要预处理必胜必败局面吧?

这里我开一个搜 16 层的东西在旁边跑,跑了 20s 出来了。

给一下打表程序:

#include<bits/stdc++.h>
#define write cout<<"player:\n"<<pl[0]<<' '<<pl[1]<<"\ncomputer:\n"<<cp[0]<<' '<<cp[1]<<'\n'<<'\n'
using namespace std;
const int w=16;
int dp[10][10][10][10];
int lx(int a[2])
{
    int b[2];
    b[0]=a[0],b[1]=a[1];
    if(b[0]>b[1]) swap(b[0],b[1]);
    if(b[0]==3&&b[1]==3||b[0]==3&&b[1]==5||b[0]==1&&b[1]==6) return 1;
    if(b[0]==1&&b[1]==9) return 2;
    if(b[0]==1&&b[1]==5) return -1;
    if(b[0]==5&&b[1]==5) return -2;
    return 0;
}
int jm(int a[2],int b[2])
{
    if(lx(a)<=0) return 0;
    if(lx(a)+lx(b)>0) return 1;
    if(lx(a)==1)
    {
        if(lx(b)==-1) b[0]=b[1]=1;
        else b[0]=5,b[1]=1;
    }
    else b[0]=b[1]=1;
    return 0;
}
int dp2[10][10][10][10][10];
int dfs(int c,int a[2],int b[2],int f)
{
    int ca[2],cb[2],i,j,sum=0;
    ca[0]=a[0],ca[1]=a[1],cb[0]=b[0],cb[1]=b[1];
    if(c==w) return 0;
    if(dp2[c][a[0]][a[1]][b[0]][b[1]]!=-1) return dp2[c][a[0]][a[1]][b[0]][b[1]];
    if(c%2==0) 
    {
        for(i=0;i<2;i++)
            for(j=0;j<2;j++)
            {
                cb[i]=(cb[i]+ca[j])%10;
                if(dp[ca[0]][ca[1]][cb[0]][cb[1]]) return dp2[c][a[0]][a[1]][b[0]][b[1]]=0;
                ca[0]=a[0],ca[1]=a[1],cb[0]=b[0],cb[1]=b[1];
            }
        for(i=0;i<2;i++)
            for(j=0;j<2;j++)
            {
                cb[i]=(cb[i]+ca[j])%10;
                jm(cb,ca);
                sum+=dfs(c+1,ca,cb,f/4);
                ca[0]=a[0],ca[1]=a[1],cb[0]=b[0],cb[1]=b[1];
            }
    }
    else
    {
        for(i=0;i<2;i++)
            for(j=0;j<2;j++)
            {
                ca[i]=(ca[i]+cb[j])%10;
                if(dp[cb[0]][cb[1]][ca[0]][ca[1]]) return dp2[c][a[0]][a[1]][b[0]][b[1]]=f; 
                ca[0]=a[0],ca[1]=a[1],cb[0]=b[0],cb[1]=b[1];
            }
        for(i=0;i<2;i++)
            for(j=0;j<2;j++)
            {
                ca[i]=(ca[i]+cb[j])%10;
                jm(ca,cb);
                sum+=dfs(c+1,ca,cb,f/4);
                ca[0]=a[0],ca[1]=a[1],cb[0]=b[0],cb[1]=b[1];
            }
    }
    return dp2[c][a[0]][a[1]][b[0]][b[1]]=sum;
}
bool ask(int a[2],int b[2],int deep)
{
    if(deep%2&&jm(b,a)) return 1;
    if(deep%2==0&&jm(a,b)) return 0;
    if(deep>w) return 0;
    int ca[2],cb[2],i,j;
    if(deep%2)
    {
        for(i=0;i<2;i++)
            for(j=0;j<2;j++)
            {
                ca[0]=a[0],ca[1]=a[1],cb[0]=b[0],cb[1]=b[1];
                ca[i]=(ca[i]+cb[j])%10;
                if(ask(ca,cb,deep+1)==0) return 0;
            }
        return 1;
    }
    else
    {
        for(i=0;i<2;i++)
            for(j=0;j<2;j++)
            {
                ca[0]=a[0],ca[1]=a[1],cb[0]=b[0],cb[1]=b[1];
                cb[i]=(cb[i]+ca[j])%10;
               if(ask(ca,cb,deep+1)==1) return 1;   
            }
        return 0;
    }
}
void jfca(int a[2],int b[2])
{
    int ca[2],cb[2],s=0,mx=-1073741825,x,y,i,j,sum;
    for(i=0;i<2;i++)
        for(j=0;j<2;j++)
        {
            ca[0]=a[0],ca[1]=a[1],cb[0]=b[0],cb[1]=b[1];
            ca[i]=(ca[i]+cb[j])%10;
            if(jm(ca,cb))
            {
                a[i]=(a[i]+b[j])%10;
                cout<<i<<' '<<j<<endl;
                return;
            }       
        }
    for(i=0;i<2;i++)
        for(j=0;j<2;j++)
        {
            ca[0]=a[0],ca[1]=a[1],cb[0]=b[0],cb[1]=b[1];
            ca[i]=(ca[i]+cb[j])%10;
            jm(ca,cb);
            sum=dfs(0,ca,cb,1073741824);
            if(sum>mx)
            {
                mx=sum;
                x=i,y=j;
            }
        }
    cout<<x<<' '<<y<<endl;
    a[x]=(a[x]+b[y])%10;
}
void init()
{
    memset(dp2,-1,sizeof(dp2));
    int i,j,k,l,a[2],b[2],sum=0;
    for(i=0;i<10;i++)
        for(j=0;j<10;j++)
            for(k=0;k<10;k++)
                for(l=0;l<10;l++)
                {
                    a[0]=i;
                    a[1]=j;
                    b[0]=k;
                    b[1]=l;
                    dp[i][j][k][l]=ask(a,b,1); 
                    cout<<dp[i][j][k][l]<<',';
                } 
}
int main()
{
    freopen("out.txt","w",stdout);
    init();
}

然后跑到了七十多 ms。

然后发现了一件神奇的事情:这题可以打表!太牛逼了!!1后知后觉

对于每个状态给出反应措施。

#include<bits/stdc++.h>
#define write cout<<"player:\n"<<pl[0]<<' '<<pl[1]<<"\ncomputer:\n"<<cp[0]<<' '<<cp[1]<<'\n'<<'\n'
using namespace std;
int w=10;
int 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
int lx(int a[2])
{
    int b[2];
    b[0]=a[0],b[1]=a[1];
    if(b[0]>b[1]) swap(b[0],b[1]);
    if(b[0]==3&&b[1]==3||b[0]==3&&b[1]==5||b[0]==1&&b[1]==6) return 1;
    if(b[0]==1&&b[1]==9) return 2;
    if(b[0]==1&&b[1]==5) return -1;
    if(b[0]==5&&b[1]==5) return -2;
    return 0;
}
int jm(int a[2],int b[2])
{
    if(lx(a)<=0) return 0;
    if(lx(a)+lx(b)>0) return 1;
    if(lx(a)==1)
    {
        if(lx(b)==-1) b[0]=b[1]=1;
        else b[0]=5,b[1]=1;
    }
    else b[0]=b[1]=1;
    return 0;
}
int dp2[10][10][10][10][10];
int dfs(int c,int a[2],int b[2],int f)
{
    int ca[2],cb[2],i,j,sum=0;
    ca[0]=a[0],ca[1]=a[1],cb[0]=b[0],cb[1]=b[1];
    if(c==w) return 0;
    if(dp2[c][a[0]][a[1]][b[0]][b[1]]!=-1) return dp2[c][a[0]][a[1]][b[0]][b[1]];
    if(c%2==0) 
    {
        for(i=0;i<2;i++)
            for(j=0;j<2;j++)
            {
                cb[i]=(cb[i]+ca[j])%10;
                if(dp[ca[0]][ca[1]][cb[0]][cb[1]]) return dp2[c][a[0]][a[1]][b[0]][b[1]]=0;
                ca[0]=a[0],ca[1]=a[1],cb[0]=b[0],cb[1]=b[1];
            }
        for(i=0;i<2;i++)
            for(j=0;j<2;j++)
            {
                cb[i]=(cb[i]+ca[j])%10;
                jm(cb,ca);
                sum+=dfs(c+1,ca,cb,f/4);
                ca[0]=a[0],ca[1]=a[1],cb[0]=b[0],cb[1]=b[1];
            }
    }
    else
    {
        for(i=0;i<2;i++)
            for(j=0;j<2;j++)
            {
                ca[i]=(ca[i]+cb[j])%10;
                if(dp[cb[0]][cb[1]][ca[0]][ca[1]]) return dp2[c][a[0]][a[1]][b[0]][b[1]]=f; 
                ca[0]=a[0],ca[1]=a[1],cb[0]=b[0],cb[1]=b[1];
            }
        for(i=0;i<2;i++)
            for(j=0;j<2;j++)
            {
                ca[i]=(ca[i]+cb[j])%10;
                jm(ca,cb);
                sum+=dfs(c+1,ca,cb,f/4);
                ca[0]=a[0],ca[1]=a[1],cb[0]=b[0],cb[1]=b[1];
            }
    }
    return dp2[c][a[0]][a[1]][b[0]][b[1]]=sum;
}
pair<int,int> jfca(int a[2],int b[2])
{
    int ca[2],cb[2],s=0,mx=-1073741825,x,y,i,j,sum;
    for(i=0;i<2;i++)
        for(j=0;j<2;j++)
        {
            ca[0]=a[0],ca[1]=a[1],cb[0]=b[0],cb[1]=b[1];
            ca[i]=(ca[i]+cb[j])%10;
            if(jm(ca,cb))
            {
                a[i]=(a[i]+b[j])%10;
                return make_pair(i,j);
            }       
        }
    for(i=0;i<2;i++)
        for(j=0;j<2;j++)
        {
            ca[0]=a[0],ca[1]=a[1],cb[0]=b[0],cb[1]=b[1];
            ca[i]=(ca[i]+cb[j])%10;
            jm(ca,cb);
            sum=dfs(0,ca,cb,1073741824);
            if(sum>mx)
            {
                mx=sum;
                x=i,y=j;
            }
        }
    a[x]=(a[x]+b[y])%10;
    return make_pair(x,y);
}
int main()
{
    freopen("out2.txt","w",stdout);
    memset(dp2,-1,sizeof(dp2));
    int i,j,k,l,a[2],b[2],sum=0;
    pair<int,int> ans;
    for(i=0;i<10;i++)
        for(j=0;j<10;j++)
            for(k=0;k<10;k++)
                for(l=0;l<10;l++)
                {
                    a[0]=i;
                    a[1]=j;
                    b[0]=k;
                    b[1]=l;
                    ans=jfca(a,b);
                    cout<<ans.first<<',';
                } 
    freopen("out3.txt","w",stdout);
    for(i=0;i<10;i++)
        for(j=0;j<10;j++)
            for(k=0;k<10;k++)
                for(l=0;l<10;l++)
                {
                    a[0]=i;
                    a[1]=j;
                    b[0]=k;
                    b[1]=l;
                    ans=jfca(a,b);
                    cout<<ans.second<<',';
                } 
}

开十层放心大胆的跑,很快就出来了。

然后一交,47 行,跑的飞快。