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=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //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[][]={{-,,-,},{-,,,-},{,-,,-},{,-,-,}};
 int ma[N][N],f[N][N][N][N],xy[*N][];
 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=;i<n;i++){
             for(j=;j<n;j++)
                 scanf("%d",&ma[i][j]);
         }
         mem(f,);
         f[][][][]=ma[][]+ma[][]+ma[][];
         up=(n-)<<|;x0=,y0=;w=;
         for(i=;i<up-;i++){
             i<=up/?(x0++,w++):(y0++,w--);
             for(j=;j<w;j++){
                 xy[j][]=x0-j;
                 xy[j][]=y0+j;
             }
             for(q=;q<w;q++){
                 for(p=q+;p<w;p++){
                     x1=xy[q][],y1=xy[q][];
                     x2=xy[p][],y2=xy[p][];
                     for(k=;k<;k++){
                         nx1=x1+d[k][],ny1=y1+d[k][];
                         nx2=x2+d[k][],ny2=y2+d[k][];
                         if(nx1==nx2 && ny1==ny2)continue;
                         if(nx1<||ny1< || nx2<||ny2<)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\n",f[n][n-][n-][n]+ma[n][n]);

     }
     return ;
 }