1036 - A Refining Company

1036 - A Refining Company

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Time Limit: 3 second(s)

Memory Limit: 32 MB

Its year 2200, planet Earth is out of resources and people are relying on the resources from other planets. There are several Refining Companies who collect these resources from other planets and bring them back to Earth. The task may sound simple, but in reality it's a challenging job. The resources are scattered and after collecting them, they have to be taken to a place where they can be refined. Since some minerals are extremely dangerous, the whole process should be done very carefully. A single tiny mistake can cause a massive explosion resulting in a huge loss.

You work in such a company who collects Uranium and Radium from planet Krypton. These minerals are used for generating powers. For simplicity you have divided planet Krypton into cells that form a matrix of m rows and n columns, where the rows go from east to west and the columns go from north to south. Your advanced mine detector has detected the approximate amount of Radium and Uranium in each cell. Your company has built two refining factories, one in West and the other in North. The factory in North is used to refine Radium and the factory in West is used to refine Uranium. 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 Radium mined at a cell can be transported to the Radium 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 belt will 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 Radium transported to the Uranium 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 Radium transported to the Radium refinery and the amount of Uranium to the Uranium refinery.

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case begins with a blank line and two integers: m - the number of rows, and n - the number columns (1 ≤ m, n ≤ 500). The next m lines describe the amount of Uranium that can be found in the cells. Each of these m lines contains n 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 m lines describe in a similar fashion the amount of Radium found in the cells. Data set is huge, so use faster i/o methods.

Output

For each case of input you have to print the case number and the maximum amount of minerals you can collect.

Sample Input	Output for Sample Input
2 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 2 3 5 10 34 0 0 0 0 0 0 50 0 0	Case 1: 98 Case 2: 50

Note

Dataset is huge. Use faster I/O methods.

题意：给你两种方案，一种是从右往左的，一种是从下往上的，然后如果在某个点是向上的那么要一直到顶端，同理向左的，并且这写方案不能交叉，问最大能运多少。

dp[i][j]标是在大小为前i行前j列的最大能运输的，我们先处理所给的两个矩阵，第一个是右到左，我们处理没一行的前缀和，另一个是下到上，我们处理每一列的前缀和。

方程： dp[i][j] = max(dp[i-1][j] + __a[i][j],dp[i][j-1] + __b[i][j]);表示当前的位置要么向上要么向左。

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<stdlib.h>
 4 #include<iostream>
 5 #include<math.h>
 6 #include<string.h>
 7 #include<queue>
 8 using namespace std;
 9 typedef long long LL;
10 int a[600][600];
11 int b[600][600];
12 int dp[600][600];
13 int __a[600][600];
14 int __b[600][600];
15 int main(void)
16 {
17         int T;
18         scanf("%d",&T);
19         int __ca=0;
20         while(T--)
21         {
22                 __ca++;
23                 int n,m;
24                 scanf("%d %d",&n,&m);
25                 int i,j ;
26                 for(i =  1; i <= n; i++)
27                 {
28                         for(j = 1; j <= m; j++)
29                         {
30                                 scanf("%d",&a[i][j]);
31                         }
32                 }
33                 for(i = 1; i <=n ; i++)
34                 {
35                         for(j = 1; j <= m; j++)
36                         {
37                                 scanf("%d",&b[i][j]);
38                         }
39                 }
40                 memset(__a,0,sizeof(__a));
41                 memset(__b,0,sizeof(__b));
42                 memset(dp,0,sizeof(dp));
43                 for(i = 1; i <= n;i++)
44                 {
45                         for(j = 1; j <=m; j++)
46                                 __a[i][j]=__a[i][j-1]+a[i][j];
47                 }
48                 for( i = 1; i <= m; i++)
49                 {
50                         for(j = 1; j <= n; j++)
51                                 __b[j][i]=__b[j-1][i]+b[j][i];
52                 }int maxx=0;
53                 for(i = 1; i <= n ; i++)
54                 {
55                         for( j = 1; j <=m; j++)
56                         {
57                                dp[i][j] = max(dp[i-1][j] + __a[i][j],dp[i][j-1] + __b[i][j]);
58                         }
59                 }
60                 printf("Case %d: %d\n",__ca,dp[n][m]);
61         }return 0;
62 }