题目链接：http://codeforces.com/problemset/problem/430/D
题意：两个人在操场上分别从左上角和左下角到右下角和右上角，左上角的人只能向右向下，左下角的人只能向右向上。他们必须在某一时刻见面一次，操场上每个点有一个值，问如何选取交点使得他们的值总和最大，他们交点的值不计。
先在四个角到对角线的角做dp，求出最长的路径，再根据两个人的行走情况枚举所有可能的交点，并统计最值。
1 /* 2 ━━━━━┒ギリギリ♂ eye！ 3 ┓┏┓┏┓┃キリキリ♂ mind！ 4 ┛┗┛┗┛┃＼○／ 5 ┓┏┓┏┓┃ / 6 ┛┗┛┗┛┃ノ) 7 ┓┏┓┏┓┃ 8 ┛┗┛┗┛┃ 9 ┓┏┓┏┓┃ 10 ┛┗┛┗┛┃ 11 ┓┏┓┏┓┃ 12 ┛┗┛┗┛┃ 13 ┓┏┓┏┓┃ 14 ┃┃┃┃┃┃ 15 ┻┻┻┻┻┻ 16 */ 17 #include <algorithm> 18 #include <iostream> 19 #include <iomanip> 20 #include <cstring> 21 #include <climits> 22 #include <complex> 23 #include <fstream> 24 #include <cassert> 25 #include <cstdio> 26 #include <bitset> 27 #include <vector> 28 #include <deque> 29 #include <queue> 30 #include <stack> 31 #include <ctime> 32 #include <set> 33 #include <map> 34 #include <cmath> 35 using namespace std; 36 #define fr first 37 #define sc second 38 #define cl clear 39 #define BUG puts("here!!!") 40 #define W(a) while(a--) 41 #define pb(a) push_back(a) 42 #define Rint(a) scanf("%d", &a) 43 #define Rll(a) scanf("%I64d", &a) 44 #define Rs(a) scanf("%s", a) 45 #define Cin(a) cin >> a 46 #define FRead() freopen("in", "r", stdin) 47 #define FWrite() freopen("out", "w", stdout) 48 #define Rep(i, len) for(int i = 0; i < (len); i++) 49 #define For(i, a, len) for(int i = (a); i < (len); i++) 50 #define Cls(a) memset((a), 0, sizeof(a)) 51 #define Clr(a, x) memset((a), (x), sizeof(a)) 52 #define Full(a) memset((a), 0x7f7f7f, sizeof(a)) 53 #define lrt rt << 1 54 #define rrt rt << 1 | 1 55 #define pi 3.14159265359 56 #define RT return 57 #define lowbit(x) x & (-x) 58 #define onecnt(x) __builtin_popcount(x) 59 typedef long long LL; 60 typedef long double LD; 61 typedef unsigned long long ULL; 62 typedef pair<int, int> pii; 63 typedef pair<string, int> psi; 64 typedef pair<LL, LL> pll; 65 typedef map<string, int> msi; 66 typedef vector<int> vi; 67 typedef vector<LL> vl; 68 typedef vector<vl> vvl; 69 typedef vector<bool> vb; 70 71 const int maxn = 1010; 72 int dp[5][maxn][maxn]; 73 int a[maxn][maxn]; 74 int n, m; 75 76 int main() { 77 // FRead(); 78 scanf("%d %d", &n, &m); 79 for (int i = 1; i <= n; ++i) 80 for (int j = 1; j <= m; ++j) 81 scanf("%d", &a[i][j]); 82 for (int i = 1; i <= n; ++i) 83 for (int j = 1; j <= m; ++j) 84 dp[1][i][j] = a[i][j] + max(dp[1][i-1][j], dp[1][i][j-1]); 85 for (int i = n; i >= 1; --i) 86 for (int j = m; j >= 1; --j) 87 dp[4][i][j] = a[i][j] + max(dp[4][i][j+1], dp[4][i+1][j]); 88 for (int i = n; i >= 1; --i) 89 for (int j = 1; j <= m; ++j) 90 dp[3][i][j] = a[i][j] + max(dp[3][i][j-1], dp[3][i+1][j]); 91 for (int i = 1; i <= n; ++i) 92 for (int j = m; j >= 1; --j) 93 dp[2][i][j] = a[i][j] + max(dp[2][i-1][j], dp[2][i][j+1]); 94 int ans = 0; 95 for (int i = 2; i < n; ++i) 96 for (int j = 2; j < m; ++j) { 97 ans = max(ans, dp[1][i][j-1] + dp[4][i][j+1] + dp[3][i+1][j] + dp[2][i-1][j]); 98 ans = max(ans, dp[1][i-1][j] + dp[4][i+1][j] + dp[3][i][j-1] + dp[2][i][j+1]); 99 } 100 printf("%d ", ans); 101 return 0; 102 }