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HDU 2845 Beans (DP)

时间:2014-12-13 19:17:25      阅读:284      评论:0      收藏:0      [点我收藏+]

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Beans
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Description

Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled with different qualities beans. Meantime, there is only one bean in any 1*1 grid. Now you want to eat the beans and collect the qualities, but everyone must obey by the following rules: if you eat the bean at the coordinate(x, y), you can’t eat the beans anyway at the coordinates listed (if exiting): (x, y-1), (x, y+1), and the both rows whose abscissas are x-1 and x+1. 
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Now, how much qualities can you eat and then get ?
 

Input

There are a few cases. In each case, there are two integer M (row number) and N (column number). The next M lines each contain N integers, representing the qualities of the beans. We can make sure that the quality of bean isn‘t beyond 1000, and 1<=M*N<=200000.
 

Output

For each case, you just output the MAX qualities you can eat and then get.
 

Sample Input

4 6 11 0 7 5 13 9 78 4 81 6 22 4 1 40 9 34 16 10 11 22 0 33 39 6
 

Sample Output

242
 

 

 思路:
STEP_1:先以行为单位来算,设DP数组存的是以此位置为起点能得到的最大值,在此前提下,DP[i][j]因为相邻的不能走,所以要么加上DP[i][j + 2],要么加上DP[i][j + 3] (假设不越界),取最大的那个就好。从右往左依次计算这一行每个元素的DP值,记录下最大的,放到DP[i][0]里。
STEP_2:重复第一步,只不过对象换成了DP数组里每行0号单元的值,因为此单元放的是此行内的最大值。从上往下,每次选定一行后设此行为最终答案里最下面那行,因为选定一行之后它的上面和下面那一行就废掉了,所以每一行要么加上它上面第二行,要么加上它上面第三行。同理,记录下最大值,最大值即答案。(实际上答案不是最后一行就是倒数第二行,因为没有负数,只会越加越大,不过只有1行的时候就会越界,虽然HDU上的数据貌似没1行的)。
 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 #include<string.h>
 4 #define    MAX    200105
 5 
 6 int    main(void)
 7 {
 8     int    n,m;
 9     int    max,temp_1,temp_2,ans;
10 
11     while(scanf("%d%d",&n,&m) != EOF)
12     {
13         int    dp[n + 10][m + 10];
14 
15         memset(dp,0,sizeof(dp));
16         for(int i = 1;i <= n;i ++)
17             for(int j = 1;j <= m;j ++)
18                 scanf("%d",&dp[i][j]);
19 
20         for(int i = 1;i <= n;i ++)
21         {
22             max = dp[i][m];
23             for(int j = m;j >= 1;j --)
24             {
25                 dp[i][j] += dp[i][j + 2] > dp[i][j + 3] ? dp[i][j + 2] : dp[i][j + 3];
26                 max = max > dp[i][j] ? max : dp[i][j];
27             }
28             dp[i][0] = max;
29         }
30 
31         max = dp[1][0];
32         dp[3][0] += dp[1][0];
33         for(int i = 4;i <= n;i ++)
34         {
35             dp[i][0] += dp[i - 2][0] > dp[i - 3][0] ? dp[i - 2][0] : dp[i - 3][0];
36             max = max > dp[i][0] ? max : dp[i][0];
37         }
38 
39         printf("%d\n",max);
40     }
41 
42     return    0;
43 }

 

 

HDU 2845 Beans (DP)

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原文地址:http://www.cnblogs.com/xz816111/p/4161687.html

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