hdu-Beans(动态规划，nyoj-234-吃土豆)

Beans

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2168 Accepted Submission(s): 1094

Problem 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.

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

thebeans. 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

Source

2009 Multi-University Training Contest 4 - Host by HDU

Recommend

gaojie

nyoj-234吃土豆

一样的题目，测试数据不一样多而已


/*把二维按行列分解然后dp。
 
dp[i][0] 表示不取第i个能到的最大值
 
所以dp[i][0] = Max(dp[i-1][0],dp[i-1][1]);
 
dp[i][1] 表示取第i个能到的最大值
 
所以dp[i][1] = dp[i-1][0]+value[i];
 
row[i]记录下每一行的最优情况,然后再按照上面的方法进行dp*/
#include<stdio.h>
  
#define  Maxn 200001 
#define  Max(a,b)  (a>b?a:b)
int dp[Maxn][2];
int value[Maxn];
int row[Maxn];
int M,N;
int main()
{//Accepted 2845 359MS 236K 835 B G++ 1983210400 
 int i,j,k;
 while(~scanf("%d %d",&M,&N))
 {
    dp[0][0]=dp[0][1]=0;
     for(i=1;i<=M;++i) //把n,m都扩大2，方便dp
     { 
         for(j=1;j<=N;++j)
         {
             scanf("%d",&value[j]);
             
             dp[j][0] = Max(dp[j-1][0],dp[j-1][1]); //累积i行到j列的最大和
             dp[j][1] = dp[j-1][0]+value[j];
         }
         row[i] = Max(dp[N][0],dp[N][1]);
         
     }
        dp[0][0] = 0;dp[0][1] = 0;
       
	    for (i = 1; i <= M; i++)
        {
            dp[i][0] = Max(dp[i-1][0],dp[i-1][1]);
            dp[i][1] = dp[i-1][0]+row[i];
        }
        printf("%d\n",Max(dp[M][1],dp[M][0]));
    }
 
 return 0;
}