HDU-2686 Matrix 多进程DP

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2686

　　经典的多进程DP，比较简单。f[x1][y1][x2][y2]表示起点到点(x1,y1)和(x2,y2)的最优值，然后分层转移就可以了，每一层为斜向右的线。。

//STATUS:C++_AC_46MS_6172KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef long long LL;
typedef unsigned long long ULL;
//const
const int N=35;
const int INF=0x3f3f3f3f;
const int MOD=1000000007,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

int d[4][4]={{-1,0,-1,0},{-1,0,0,-1},{0,-1,0,-1},{0,-1,-1,0}};
int ma[N][N],f[N][N][N][N],xy[2*N][2];
int n;

int main(){
 //   freopen("in.txt","r",stdin);
    int i,j,q,p,k,x0,y0,w,x1,y1,x2,y2,nx1,ny1,nx2,ny2,up;
    while(~scanf("%d",&n))
    {
        for(i=0;i<n;i++){
            for(j=0;j<n;j++)
                scanf("%d",&ma[i][j]);
        }
        mem(f,0);
        f[1][0][0][1]=ma[0][0]+ma[1][0]+ma[0][1];
        up=(n-1)<<1|1;x0=1,y0=0;w=2;
        for(i=2;i<up-1;i++){
            i<=up/2?(x0++,w++):(y0++,w--);
            for(j=0;j<w;j++){
                xy[j][0]=x0-j;
                xy[j][1]=y0+j;
            }
            for(q=0;q<w;q++){
                for(p=q+1;p<w;p++){
                    x1=xy[q][0],y1=xy[q][1];
                    x2=xy[p][0],y2=xy[p][1];
                    for(k=0;k<4;k++){
                        nx1=x1+d[k][0],ny1=y1+d[k][1];
                        nx2=x2+d[k][2],ny2=y2+d[k][3];
                        if(nx1==nx2 && ny1==ny2)continue;
                        if(nx1<0||ny1<0 || nx2<0||ny2<0)continue;
                        f[x1][y1][x2][y2]=Max(f[x1][y1][x2][y2],f[nx1][ny1][nx2][ny2]+ma[x1][y1]+ma[x2][y2]);
                    }
                }
            }
        }

        n--;
        printf("%d
",f[n][n-1][n-1][n]+ma[n][n]);

    }
    return 0;
}