Codeforces 429B Working out

Test instructions: Matrix A line from top left to bottom right, a top right to bottom left, there can only be one intersection, and the intersection value is not counted, ask you how much value you can get.

Problem Solving ideas: First find out four matrices, and then enumerate the points and the way to enter the point, where the DP problem requires special consideration of the boundary.

Problem Solving Code:

1 //File name:429b.cpp2 //Author:darkdream3 //Created time:2015 March 08 Sunday 10:13 53 seconds4 5#include <vector>6#include <list>7#include <map>8#include <Set>9#include <deque>Ten#include <stack> One#include <bitset> A#include <algorithm> -#include <functional> -#include <numeric> the#include <utility> -#include <sstream> -#include <iostream> -#include <iomanip> +#include <cstdio> -#include <cmath> +#include <cstdlib> A#include <cstring> at#include <ctime> - #defineLL Long Long - using namespacestd; - inta[1005][1005]; - intldp[1005][1005]; - intrdp[1005][1005]; in intudp[1005][1005]; - intddp[1005][1005]; to intMain () { +   intN, M; -scanf"%d%d",&n,&m); the    for(inti =1; I <= N;i + +) *        for(intj =1; J <= M;j + +) $scanf"%d",&a[i][j]);Panax Notoginseng    for(inti =1; I <= N;i + +) -   { the       for(intj =1; J <= M;j + +) +      { ALDP[I][J] = A[i][j] + max (ldp[i][j-1],ldp[i-1][j]); the      } +   } -    for(inti = N;i >=1; I--) $   { $       for(intj =1; J <= M;j + +) -      { -RDP[I][J] = A[i][j] + max (rdp[i][j-1],rdp[i+1][j]); the      } -   }Wuyi    for(inti = N;i >=1; I--) the   { -       for(intj = M;j >=1; J--) Wu      { -UDP[I][J] = A[i][j] + max (udp[i][j+1],udp[i+1][j]); About      } $   } -    for(inti =1; I <= N;i + +) -   { -       for(intj = m; J >=1; J--) A      { +DDP[I][J] = A[i][j] + max (ddp[i][j+1],ddp[i-1][j]); the      } -   } $   intsum =0 ;  the    for(inti =2; i < N;i + +) the        for(intj =2; J < M;j + +) the       { the           intTsum =sum; -sum = max (sum,ldp[i-1][J] + udp[i+1][J] + rdp[i][j-1] + ddp[i][j+1]) ;  insum = max (sum,rdp[i+1][J] + ddp[i-1][J] + ldp[i][j-1] + udp[i][j+1]) ;  the       } theprintf"%d\n", sum); About return 0; the}

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Codeforces 429B Working out