「Luogu4363/BZOJ5248」[九省联考2018]一双木棋chess

学校省选模拟居然拿九省联考来考

然而我还是\(too\space young\)，搞不懂什么叫最优

让二者的答案最接近可以拿到\(25\)分的好成绩

problem

Solution

首先可以知道菲菲想要最大化\(ans=Ansa-Ansb\)，牛牛想要最小化

那么我们可以用对抗搜索大力爆搜

可以拿到50分

#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long ll;template <typename T> void read(T &t)
{t=0;int f=0;char c=getchar();while(!isdigit(c)){f|=c=='-';c=getchar();}while(isdigit(c)){t=t*10+c-'0';c=getchar();}if(f)t=-t;
}const int maxn=15,maxm=15;
int n,m;
int A[maxn][maxm],B[maxn][maxm];
int ocr[maxn][maxm];int dfs(int step)
{if(step==n*m+1)return 0;int re=0;if(step&1)re=-0x3f3f3f3f;else re=0x3f3f3f3f;for(register int i=1;i<=n;++i)for(register int j=1;j<=m;++j)if(!ocr[i][j] && ocr[i-1][j] && ocr[i][j-1]){ocr[i][j]=1;if(step&1)re=max(re,dfs(step+1)+A[i][j]);else re=min(re,dfs(step+1)-B[i][j]);ocr[i][j]=0;}return re;
}int main()
{read(n);read(m);for(register int i=1;i<=n;++i)for(register int j=1;j<=m;++j)read(A[i][j]);for(register int i=1;i<=n;++i)for(register int j=1;j<=m;++j)read(B[i][j]);for(register int i=0;i<=n;++i)ocr[i][0]=1;for(register int i=0;i<=m;++i)ocr[0][i]=1;printf("%d",dfs(1));return 0;
}

显然这个东西可以记忆化一下

用\(map\)存一下棋盘的哈希值，吸氧的情况下是能A掉的

#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <map>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;template <typename T> void read(T &t)
{t=0;int f=0;char c=getchar();while(!isdigit(c)){f|=c=='-';c=getchar();}while(isdigit(c)){t=t*10+c-'0';c=getchar();}if(f)t=-t;
}const int maxn=15,maxm=15;int n,m;
int A[maxn][maxm],B[maxn][maxm];
int ocr[maxn][maxm];
map<ull,int> rec;ull Hash()
{ull re=0;for(register int i=1;i<=n;++i)for(register int j=1;j<=m;++j)re=re*3ull+ocr[i][j];return re;
}int dfs(int step)
{if(step==n*m+1)return 0;int re=0;ull h=Hash();if(rec.find(h)!=rec.end())return rec[h];if(step&1)re=-0x3f3f3f3f;else re=0x3f3f3f3f;for(register int i=1;i<=n;++i)for(register int j=1;j<=m;++j)if(!ocr[i][j] && ocr[i-1][j] && ocr[i][j-1]){ocr[i][j]=1;if(step&1)re=max(re,dfs(step+1)+A[i][j]);else re=min(re,dfs(step+1)-B[i][j]);ocr[i][j]=0;}return rec[h]=re;
}int main()
{read(n);read(m);for(register int i=1;i<=n;++i)for(register int j=1;j<=m;++j)read(A[i][j]);for(register int i=1;i<=n;++i)for(register int j=1;j<=m;++j)read(B[i][j]);for(register int i=0;i<=n;++i)ocr[i][0]=1;for(register int i=0;i<=m;++i)ocr[0][i]=1;printf("%d",dfs(1));return 0;
}

经过百度一下之后，发现此类对抗搜索还有一种优化，叫做\(Alpha-Beta\)优化

在此仅放上介绍链接，不再赘述

对于此题，我们如果使用\(Alpha-Beta\)优化，也能获得70分的成绩

#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;template <typename T> void read(T &t)
{t=0;int f=0;char c=getchar();while(!isdigit(c)){f|=c=='-';c=getchar();}while(isdigit(c)){t=t*10+c-'0';c=getchar();}if(f)t=-t;
}const int maxn=15,maxm=15;int n,m;
int A[maxn][maxm],B[maxn][maxm];
int ocr[maxn][maxm];const int inf=0x3f3f3f3f;int dfs(int step,int alpha,int beta,int nowa,int nowb)
{if(step==n*m+1)return nowa-nowb;if(step&1){for(register int i=1;i<=n;++i)for(register int j=1;j<=m;++j)if(!ocr[i][j] && ocr[i-1][j] && ocr[i][j-1]){ocr[i][j]=1;alpha=max(alpha,dfs(step+1,alpha,beta,nowa+A[i][j],nowb));ocr[i][j]=0;if(alpha>=beta)return alpha;}return alpha;}else{for(register int i=1;i<=n;++i)for(register int j=1;j<=m;++j)if(!ocr[i][j] && ocr[i-1][j] && ocr[i][j-1]){ocr[i][j]=1;beta=min(beta,dfs(step+1,alpha,beta,nowa,nowb+B[i][j]));ocr[i][j]=0;if(alpha>=beta)return beta;}return beta;}
}int main()
{read(n);read(m);for(register int i=1;i<=n;++i)for(register int j=1;j<=m;++j)read(A[i][j]);for(register int i=1;i<=n;++i)for(register int j=1;j<=m;++j)read(B[i][j]);for(register int i=0;i<=n;++i)ocr[i][0]=1;for(register int i=0;i<=m;++i)ocr[0][i]=1;printf("%d",dfs(1,-inf,inf,0,0));return 0;
}