【NOIP2007提高组】矩阵取数游戏

本题DP+高精度即可。
首先我们可以发现它的贡献只与行有关系，于是就分成n行，每行都做DP，然后将max加起来即可。（PS：用高精度实现）
上标：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define mo 100000
#define ll long long
using namespace std;
int n,m,c[81][81];
struct gjd
{
	int a[51],top;
}ans,f[81][81],e[81],s,d;

inline int read()
{
	int x=0; char c=getchar();
	while (c<'0' || c>'9') c=getchar();
	while (c>='0' && c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x;
}

gjd cheng(gjd a,int b)
{
	int x=0;
	for (int i=1;i<=a.top;i++)
	{
		a.a[i]=a.a[i]*b+x;
		x=a.a[i]/mo;a.a[i]%=mo;
	}
	while (x) a.a[++a.top]=x%mo,x/=mo;
	return a;
}

gjd plus(gjd a,gjd b)
{
	int x=0;
	memset(d.a,0,sizeof(d.a));
	d.top=max(a.top,b.top)+1;
	for (int i=1;i<=d.top;i++)
	{
		d.a[i]=a.a[i]+b.a[i]+x;
		x=d.a[i]/mo;d.a[i]%=mo;
	}
	while (!d.a[d.top]) d.top--;
	return d;
}

gjd ma(gjd a,gjd b)
{
	if (a.top>b.top) return a;
	else if (a.top<b.top) return b;
	for (int i=a.top;i>0;i--)
		if (a.a[i]>b.a[i]) return a;
		else if (a.a[i]<b.a[i]) return b;
	return a;
}

int main()
{
	freopen("game.in","r",stdin);
	freopen("game.out","w",stdout);
	n=read(),m=read();
	e[0].a[1]=1,e[0].top=1;
	for (int i=1;i<=m;i++)
		e[i]=cheng(e[i-1],2);
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++) c[i][j]=read();
	for (int i=1;i<=n;i++)
	{
		memset(f,0,sizeof(f));
		
		f[0][m].a[1]=0,f[0][m].top=1;
		for (int j=1;j<=m;j++)
		{
			f[j][m]=ma(f[j][m],plus(f[j-1][m],cheng(e[j],c[i][j])));
			for (int k=m-1;k>=j;k--)
				f[j][k]=ma(f[j][k],ma(plus(f[j-1][k],cheng(e[m-k+j],c[i][j])),plus(f[j][k+1],cheng(e[m-k+j],c[i][k+1]))));
		}
		for (int j=m-1;j>=0;j--)
		{
			f[0][j]=ma(f[0][j],plus(f[0][j+1],cheng(e[m-j],c[i][j+1])));
			for (int k=1;k<=j;k++)
				f[k][j]=ma(f[k][j],ma(plus(f[k-1][j],cheng(e[m-j+k],c[i][k])),plus(f[k][j+1],cheng(e[m-j+k],c[i][j+1]))));
		}
		memset(s.a,0,sizeof(s.a));
		s.top=0;
		for (int j=0;j<=m;j++)
			s=ma(s,f[j][j]);
		ans=plus(ans,s);
	}
	printf("%d",ans.a[ans.top]);
	for (int i=ans.top-1;i>0;i--)
		printf("%05d",ans.a[i]);
	return 0;
}