poj 2984Martian Mining(dp)

                                                                                                     Martian Mining
Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 2611 Accepted: 1624
Description


The NASA Space Center, Houston, is less than 200 miles from San Antonio, Texas (the site of the ACM Finals this year). This is the place where the astronauts are trained for Mission Seven Dwarfs, the next giant leap in space exploration. The Mars Odyssey program revealed that the surface of Mars is very rich in yeyenum and bloggium. These minerals are important ingredients for certain revolutionary new medicines, but they are extremely rare on Earth. The aim of Mission Seven Dwarfs is to mine these minerals on Mars and bring them back to Earth. 


The Mars Odyssey orbiter identified a rectangular area on the surface of Mars that is rich in minerals. The area is divided into cells that form a matrix of n rows and m columns, where the rows go from east to west and the columns go from north to south. The orbiter determined the amount of yeyenum and bloggium in each cell. The astronauts will build a yeyenum refinement factory west of the rectangular area and a bloggium factory to the north. Your task is to design the conveyor belt system that will allow them to mine the largest amount of minerals. 


There are two types of conveyor belts: the first moves minerals from east to west, the second moves minerals from south to north. In each cell you can build either type of conveyor belt, but you cannot build both of them in the same cell. If two conveyor belts of the same type are next to each other, then they can be connected. For example, the bloggium mined at a cell can be transported to the bloggium refinement factory via a series of south-north conveyor belts. 


The minerals are very unstable, thus they have to be brought to the factories on a straight path without any turns. This means that if there is a south-north conveyor belt in a cell, but the cell north of it contains an east-west conveyor belt, then any mineral transported on the south-north conveyor beltwill be lost. The minerals mined in a particular cell have to be put on a conveyor belt immediately, in the same cell (thus they cannot start the transportation in an adjacent cell). Furthermore, any bloggium transported to the yeyenum refinement factory will be lost, and vice versa. 
 
Your program has to design a conveyor belt system that maximizes the total amount of minerals mined,i.e., the sum of the amount of yeyenum transported to the yeyenum refinery and the amount of bloggium transported to the bloggium refinery.
Input


The input contains several blocks of test cases. Each case begins with a line containing two integers: the number 1 ≤ n ≤ 500 of rows, and the number 1 ≤ m ≤ 500 of columns. The next n lines describe the amount of yeyenum that can be found in the cells. Each of these n lines contains m integers. The first line corresponds to the northernmost row; the first integer of each line corresponds to the westernmost cell of the row. The integers are between 0 and 1000. The next n lines describe in a similar fashion theamount of bloggium found in the cells. 


The input is terminated by a block with n = m = 0.
Output


For each test case, you have to output a single integer on a separate line: the maximum amount of mineralsthat can be mined.
Sample Input


4 4
0 0 10 9
1 3 10 0
4 2 1 3 
1 1 20 0
10 0 0 0
1 1 1 30
0 0 5 5
5 10 10 10
0 0
Sample Output


98
Hint


Huge input file, 'scanf' recommended to avoid TLE. 
题意： 一个row*col的矩阵，每个格子内有两种矿yeyenum和bloggium，并且知道它们在每个


格子内的数量是多少。最北边有bloggium的收集站，最西边有 yeyenum 的收集站。


现在要在这些格子上面安装向北或者向西的传送带(每个格子自能装一种)。问最多能采到多少矿。


传送带只能直着走，不可弯曲，不能交叉。
思路：
dp[i][j]代表着从（0，0）到（i,j）这两点间组成的矩阵最多能收集多少矿石。
bl[i][j]代表Bloggium这种矿石如果从点（i,j）向北运能运多少。
ye[i][j]代表Yeyenum这种矿石如果从点（i,j）向西运能运多少。
这题拥有这样一个性质.由于每当一个点被选取后,如果这个点是选择往上走的话,那么
     这个点的这一列上面的点就一定会全部选取,如果这个点是往左走的话,那么该点的左
     边同一行的点也应该计算在内
     设 dp[i][j]表示子矩阵(1, 1) 到 (i, j)的最大传输值,那么有动态规划方程:
     dp[i][j] = max(bl[i][j]+dp[i][j-1], ye[i][j]+dp[i-1][j])
     其中col[i][j]表示第j列从第1行到第i行的和; row[i][j]表示第i行1到j列的和
     该方程说明了一个事实就是对于当前点是否选择是影响整个这一行或者是这一列的

     除此之外,对其他地方并无影响 

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define maxn 550
int bl[maxn][maxn],ye[maxn][maxn];
int dp[maxn][maxn];
int main()
{
    int n,m;
    while(~scanf("%d%d", &n, &m),n||m)
    {
        memset(bl, 0, sizeof(bl));
        memset(ye, 0, sizeof(ye));
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= m; j++)
            {
                scanf("%d",&ye[i][j]);
                ye[i][j] += ye[i][j-1];
            }
        }
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= m; j++)
            {
                scanf("%d",&bl[i][j]);
                bl[i][j] += bl[i-1][j];
            }
        }

        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= m; j++)
            {
                dp[i][j] = max(dp[i-1][j]+ye[i][j], dp[i][j-1]+bl[i][j]);
            }
        }
        printf("%d\n",dp[n][m]);
    }

}