递推DP UVA 1366 Martian Mining

题目传送门

 /*
     题意：抽象一点就是给两个矩阵，重叠的(就是两者选择其一)，两种铺路：从右到左和从下到上，中途不能转弯，
         到达边界后把沿途路上的权值相加求和使最大
     DP：这是道递推题，首先我题目看了老半天，看懂后写出前缀和又不知道该如何定义状态好，写不出状态转移方程，太弱了。
          dp[i][j]表示以(i, j)为右下角时求得的最大值，状态转移方程：dp[i][j] = max (dp[i-1][j] + sum1[i][j], dp[i][j-1] + sum2[i][j]);   sum1表示列的前缀，sum2表示行的前缀
 */
 /************************************************
 * Author        :Running_Time
 * Created Time  :2015-8-9 10:18:37
 * File Name     :UVA_1366.cpp
  ************************************************/

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <sstream>
 #include <cstring>
 #include <cmath>
 #include <string>
 #include <vector>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <list>
 #include <map>
 #include <set>
 #include <bitset>
 #include <cstdlib>
 #include <ctime>
 using namespace std;

 #define lson l, mid, rt << 1
 #define rson mid + 1, r, rt << 1 | 1
 typedef long long ll;
 const int MAXN = 5e2 + ;
 const int INF = 0x3f3f3f3f;
 const int MOD = 1e9 + ;
 int a[MAXN][MAXN], b[MAXN][MAXN];
 int sum1[MAXN][MAXN], sum2[MAXN][MAXN];
 int dp[MAXN][MAXN];

 int main(void)    {     //UVA 1366 Martian Mining
     int n, m;
     while (scanf ("%d%d", &n, &m) == ) {
         if (!n && !m)   break;
         memset (sum1, , sizeof (sum1));
         for (int i=; i<=n; ++i)    {
             for (int j=; j<=m; ++j)    {
                 scanf ("%d", &a[i][j]); sum1[i][j] = sum1[i][j-] + a[i][j];
             }
         }
         memset (sum2, , sizeof (sum2));
         for (int i=; i<=n; ++i)    {
             for (int j=; j<=m; ++j)    {
                 scanf ("%d", &b[i][j]); sum2[i][j] = sum2[i-][j] + b[i][j];
             }
         }
         memset (dp, , sizeof (dp));
         for (int i=; i<=n; ++i)    {
             for (int j=; j<=m; ++j)    {
                 dp[i][j] = max (dp[i-][j] + sum1[i][j], dp[i][j-] + sum2[i][j]);
             }
         }
         printf ("%d\n", dp[n][m]);
     }

     return ;
 }