C++14(g++5.4) 解法, 执行用时: 105ms, 内存消耗: 740K, 提交时间: 2019-09-15 21:35:02

#include<bits/stdc++.h>

#define FOG(x,y,z) for(register int x=y,x##_=z;x<=x##_;++x)
#define DOG(x,y,z) for(register int x=y,x##_=z;x>=x##_;--x)
#define FOR(x,y,z) for(int x=y,x##_=z;x<=x##_;++x)
#define DOR(x,y,z) for(int x=y,x##_=z;x>=x##_;--x)
#define FOR_(x,y,z,s) for(int x=y,x##_=z;x<=x##_;x+=s)
#define FOR__(x,y,z) for(int x=y,x##_=z;x<=x##_;x<<=1)
#define EOR(x,y) for(int x##_=head[x],y=edge[x##_].e;x##_;y=edge[x##_=edge[x##_].to].e)
#define EGOR(x,y,z) for(int x##_=head[x],y=edge[x##_].e,z=edge[x##_].c;x##_;y=edge[x##_=edge[x##_].to].e,z=edge[x##_].c)
#define EOD(x,y) for(int &x##_=head[x],y=edge[x##_].e;x##_;y=edge[x##_=edge[x##_].to].e)
#define SOR(x,y) for(int x=y;x;x=y&(x-1))
#define SOO(x,y) for(int x=y;;x=y&(x-1))
#define FOB(b,y) for(int b##_=y,b=bin[b##_&-b##_];b##_;b##_&=b##_-1,b=bin[b##_&-b##_])
#define While(x) for(;x;)
#define clr(x,y) memset(x,y,sizeof(x))
#define lbd(A,s,e,x) (lower_bound(A+s,A+e+1,x)-A)
#define ubd(A,s,e,x) (upper_bound(A+s,A+e+1,x)-A)
#define uni(A,x) {sort(A+1,A+x+1);x=unique(A+1,A+x+1)-A-1;}
#define uniz(A,x) {sort(A,A+x);x=unique(A,A+x)-A;}
#define szf(x) sizeof(x)
#define min3(x,y,z) min(min(x,y),z)
#define max3(x,y,z) max(max(x,y),z)

#define read2(x,y) read(x),read(y)
#define read3(x,y,z) read(x),read(y),read(z)
#define read4(x,y,z,w) read3(x,y,z),read(w)
#define reads(str) sf("%s",str)
#define readf(x) sf("%lf",&x)

#define ts (*this)
#define sf scanf
#define pf printf

#define ll long long
#define ull unsigned long long
#define uint unsigned int
#define db double
#define ct clock_t
#define ck() clock()
#define rd rand()
#define rmx RAND_MAX
#define RD T*(rd*2-rmx)


using namespace std;

template<class T>bool tomin(T &x,T y){return y<x?x=y,1:0;}
template<class T>bool tomax(T &x,T y){return x<y?x=y,1:0;}
template<class T>void read(T &x){
    char c;
    x=0;
    int f=1;
    while(c=getchar(),c<'0'||c>'9')if(c=='-')f=-1;
    do x=(x<<3)+(x<<1)+(c^48);
    while(c=getchar(),c>='0'&&c<='9');
    x*=f;
}
bool mem1;
const db Pi=acos(-1);
const int dx[]={-1,-1,0,1,1,1,0,-1};
const int dy[]={0,1,1,1,0,-1,-1,-1};
const int maxn=205;
int n;
int A[maxn][maxn];
int mark[maxn][maxn],tot;
ll solve(int x,int y,int d){
    ++tot;
    ll res=0,mx=A[x][y];
    while(mark[x][y]!=tot){
        mark[x][y]=tot;
        tomax(mx,res+=A[x][y]);
        x=(x+dx[d]+n)%n;
        y=(y+dy[d]+n)%n;
    }return mx;
}
bool mem2;
int main(){
    // cerr<<(&mem2-&mem1)/1024.0/1024<<endl;
    srand(time(NULL));
    read(n);
    FOR(i,0,n-1)FOR(j,0,n-1)read(A[i][j]);
    ll ans=-1e18;
    FOR(i,0,n-1)FOR(j,0,n-1)FOR(d,0,7)tomax(ans,solve(i,j,d));
    pf("%lld",ans);
    return 0;
}

C++11(clang++ 3.9) 解法, 执行用时: 19ms, 内存消耗: 612K, 提交时间: 2019-09-14 11:49:49

#include <bits/stdc++.h>
using namespace std;
int n,a[201][201],maxx=-10000000,i,j,k,w,x,y,z;
int main()
{
    cin>>n;
    for(i=0;i<n;i++)for(j=0;j<n;j++)cin>>a[i][j];
    for(i=0;i<n;i++)for(j=0;j<n;j++,w=0,x=0,y=0,z=0)
    for(k=0;k<n;k++)
    {
        w+=a[i][(j+k)%n];
        x+=a[(i+k)%n][j];
        y+=a[(i+k)%n][(j+k)%n];
        z+=a[(i-k+n)%n][(j+k)%n];
        maxx=max(max(max(w,x),max(y,z)),maxx);
    }
    cout<<maxx<<endl;
}