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 }
