1 条题解

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    @ 2025-8-24 22:28:53

    自动搬运

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    来自洛谷,原作者为

    avatar critnos
    咔菲好喝。

    搬运于2025-08-24 22:28:53,当前版本为作者最后更新于2021-02-16 11:33:56,作者可能在搬运后再次修改,您可在原文处查看最新版

    自动搬运只会搬运当前题目点赞数最高的题解,您可前往洛谷题解查看更多

    以下是正文


    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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0,1,0,0,0,0,0,0,0,0}; 
    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。

    • 1

    信息

    ID
    6397
    时间
    1000ms
    内存
    128MiB
    难度
    6
    标签
    递交数
    0
    已通过
    0
    上传者