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。。

  1 //STATUS:C++_AC_46MS_6172KB

  2 #include <functional>

  3 #include <algorithm>

  4 #include <iostream>

  5 //#include <ext/rope>

  6 #include <fstream>

  7 #include <sstream>

  8 #include <iomanip>

  9 #include <numeric>

 10 #include <cstring>

 11 #include <cassert>

 12 #include <cstdio>

 13 #include <string>

 14 #include <vector>

 15 #include <bitset>

 16 #include <queue>

 17 #include <stack>

 18 #include <cmath>

 19 #include <ctime>

 20 #include <list>

 21 #include <set>

 22 #include <map>

 23 using namespace std;

 24 //#pragma comment(linker,"/STACK:102400000,102400000")

 25 //using namespace __gnu_cxx;

 26 //define

 27 #define pii pair<int,int>

 28 #define mem(a,b) memset(a,b,sizeof(a))

 29 #define lson l,mid,rt<<1

 30 #define rson mid+1,r,rt<<1|1

 31 #define PI acos(-1.0)

 32 //typedef

 33 typedef long long LL;

 34 typedef unsigned long long ULL;

 35 //const

 36 const int N=35;

 37 const int INF=0x3f3f3f3f;

 38 const int MOD=1000000007,STA=8000010;

 39 const LL LNF=1LL<<60;

 40 const double EPS=1e-8;

 41 const double OO=1e15;

 42 const int dx[4]={-1,0,1,0};

 43 const int dy[4]={0,1,0,-1};

 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};

 45 //Daily Use ...

 46 inline int sign(double x){return (x>EPS)-(x<-EPS);}

 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}

 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}

 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}

 50 template<class T> inline T Min(T a,T b){return a<b?a:b;}

 51 template<class T> inline T Max(T a,T b){return a>b?a:b;}

 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}

 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}

 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}

 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}

 56 //End

 57 

 58 int d[4][4]={{-1,0,-1,0},{-1,0,0,-1},{0,-1,0,-1},{0,-1,-1,0}};

 59 int ma[N][N],f[N][N][N][N],xy[2*N][2];

 60 int n;

 61 

 62 int main(){

 63  //   freopen("in.txt","r",stdin);

 64     int i,j,q,p,k,x0,y0,w,x1,y1,x2,y2,nx1,ny1,nx2,ny2,up;

 65     while(~scanf("%d",&n))

 66     {

 67         for(i=0;i<n;i++){

 68             for(j=0;j<n;j++)

 69                 scanf("%d",&ma[i][j]);

 70         }

 71         mem(f,0);

 72         f[1][0][0][1]=ma[0][0]+ma[1][0]+ma[0][1];

 73         up=(n-1)<<1|1;x0=1,y0=0;w=2;

 74         for(i=2;i<up-1;i++){

 75             i<=up/2?(x0++,w++):(y0++,w--);

 76             for(j=0;j<w;j++){

 77                 xy[j][0]=x0-j;

 78                 xy[j][1]=y0+j;

 79             }

 80             for(q=0;q<w;q++){

 81                 for(p=q+1;p<w;p++){

 82                     x1=xy[q][0],y1=xy[q][1];

 83                     x2=xy[p][0],y2=xy[p][1];

 84                     for(k=0;k<4;k++){

 85                         nx1=x1+d[k][0],ny1=y1+d[k][1];

 86                         nx2=x2+d[k][2],ny2=y2+d[k][3];

 87                         if(nx1==nx2 && ny1==ny2)continue;

 88                         if(nx1<0||ny1<0 || nx2<0||ny2<0)continue;

 89                         f[x1][y1][x2][y2]=Max(f[x1][y1][x2][y2],f[nx1][ny1][nx2][ny2]+ma[x1][y1]+ma[x2][y2]);

 90                     }

 91                 }

 92             }

 93         }

 94 

 95         n--;

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

 97 

 98     }

 99     return 0;

100 }