题目链接：
codeforces 429B
题目大意：
给出一个矩阵，一个人从左上角走到右下角，一个人从左下角走到右上角，两个人只会在一个点相交，问两个人经过路径上的数的和最大的情况下最大和是多少。
题目分析：可以分别从四个角出发进行动态规划，dp[k][i][j]代表从第k个角出发到达i和j得到的最大的数。枚举每一个点，，然后枚举从四个角同时到大这个点的情况，只能是左上角从左侧或上侧到，右下角从右侧或下侧到，其他同理，因为只在所枚举的点相交，所以我们可以认为，左上角如果从左侧进入，因为左小也要到达这个点，所以右下角的点必须从右侧进入。同理，右上角的必须从上侧进入，如果左上角从上侧进入，同理会发现，起始只有一种情况。题目不难，但是要细心，耐心地写。 AC代码：using namespace std;int n,m;int a[MAX][MAX];int u,v;int dp[4][MAX][MAX];int main ( ){while ( ~scanf ( “%d%d” , &n , &m ) ){for ( int i = 1 ; i <= n ; i++ )for ( int j = 1 ; j <= m ; j++ )scanf ( “%d” , &a[i][j] );for ( int i = 1 ; i <= n ; i++ )for ( int j = 1 ; j <= m ; j++ ){dp[0][i][j] = 0;if ( i > 1 )dp[0][i][j] = max ( dp[0][i-1][j] , dp[0][i][j] );if ( j > 1 )dp[0][i][j] = max ( dp[0][i][j-1] , dp[0][i][j] );dp[0][i][j] += a[i][j];}for ( int i = n ; i >= 1 ; i– )for ( int j = 1 ; j <= m ; j++ ){dp[1][i][j] = 0;if ( i < n )dp[1][i][j] = max ( dp[1][i+1][j] , dp[1][i][j] );if ( j > 1 )dp[1][i][j] = max ( dp[1][i][j-1] , dp[1][i][j] );dp[1][i][j] += a[i][j];}for ( int i = 1 ; i <= n ; i++ )for ( int j = m ; j >= 1 ; j– ){dp[2][i][j] = 0;if ( i > 1 )dp[2][i][j] = max ( dp[2][i-1][j] , dp[2][i][j] );if ( j < m )dp[2][i][j] = max ( dp[2][i][j+1] , dp[2][i][j] );dp[2][i][j] += a[i][j];}for ( int i = n ; i >= 1 ; i– )for ( int j = m ; j >= 1 ; j– ){dp[3][i][j] = 0;if ( i < n )dp[3][i][j] = max ( dp[3][i+1][j] , dp[3][i][j] );if ( j < m )dp[3][i][j] = max ( dp[3][i][j+1] , dp[3][i][j] );dp[3][i][j] += a[i][j];}/*for ( int i = 1 ; i <= n ; i++ )for ( int j = 1 ; j <= m ; j++ ){dp[0][i][j+1] = max ( dp[0][i][j+1] , dp[0][i][j] + a[i][j+1] );dp[0][i+1][j] = max ( dp[0][i+1][j] , dp[0][i][j] + a[i+1][j] );}for ( int i = 1 ; i <= n ; i++ )for ( int j = m ; j >= 1; j– ){dp[2][i][j-1] = max ( dp[2][i][j-1] , dp[2][i][j] + a[i][j-1] );dp[2][i+1][j] = max ( dp[2][i+1][j] , dp[2][i][j] + a[i+1][j] );}for ( int i = n ; i >= 1 ; i– )for ( int j = 1 ; j <= m ; j++ ){dp[1][i][j+1] = max ( dp[1][i][j+1] , dp[1][i][j] + a[i][j+1] );dp[1][i-1][j] = max ( dp[1][i-1][j] , dp[1][i][j] + a[i-1][j] );}for ( int i = n ; i >= 1 ; i– )for ( int j = n ; j >= 1 ; j– ){dp[3][i][j-1] = max ( dp[3][i][j-1] , dp[3][i][j] + a[i][j-1] );dp[3][i-1][j] = max ( dp[3][i-1][j] , dp[3][i][j] + a[i-1][j] );}*/int ans = 0;for ( int i = 2; i < n ; i++ )for ( int j = 2; j < m ; j++ ){ans = max ( ans , dp[0][i-1][j]+dp[3][i+1][j]+dp[1][i][j-1]+dp[2][i][j+1] );ans = max ( ans , dp[0][i][j-1]+dp[3][i][j+1]+dp[1][i+1][j]+dp[2][i-1][j] );}printf ( “%d\n” , ans );}}
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