HDU 2845 Beans

Beans

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1819    Accepted Submission(s): 921

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 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

 

 

Source

2009 Multi-University Training Contest 4 - Host by HDU

 

 

Recommend

gaojie

 

 

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<iostream>

#include<cstdio>

#include<cstring>



using namespace std;



const int N=200010;



int m,n;

int dp[N][2],val[N],row[N];



int main(){



    //freopen("input.txt","r",stdin);



    while(~scanf("%d%d",&m,&n)){

        memset(dp,0,sizeof(dp));

        for(int i=1;i<=m;i++){

            for(int j=1;j<=n;j++){

                scanf("%d",&val[j]);

                dp[j][0]=max(dp[j-1][0],dp[j-1][1]);

                dp[j][1]=dp[j-1][0]+val[j];

            }

            row[i]=max(dp[n][0],dp[n][1]);

        }

        for(int 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][0],dp[m][1]));

    }

    return 0;

}

 

 

 

挺有意思的题。 吃了一个豆豆后，上下2行的豆豆都不能吃。这个很关键，突破口就在这。那我们只要考虑一行最多能吃多少，然后得到每行最多能吃多少。这样得到的m个数排成一行，就很之前求一行的一样了，就是如果前面一个吃了，后面的就不能再吃，也就是不能相邻。这道题和树形DP那道 不能和有直接关系的上级一起出席聚会的题很像。 所以对于一行还说，状态表示dp[i][j]，j要么取0，要么取1，表示第i个数取了没。

状态转移就是dp[i][0]=max(dp[i-1][0],dp[i-1][1])    dp[i][1]=dp[i-1][0]+val[i]

这道题只给出了M*N，不可能用2个数定一个位置。所以就用了dp[i][j]，1<=i<=M*N j=0,1

得到每行的最大值后（max(dp[i][0],dp[i][1]),i=n*k），再对这些最大值dp，方法一样，状态的第一个参数也是MAX，因为N可能为1.

 

 

#include<stdio.h>

#include<string.h>



const int maxn=200010;



int dp_row[maxn][2];

int dp_col[maxn][2];



int max(int a,int b){

    return a>b?a:b;

}



int main(){



    //freopen("input.txt","r",stdin);



    int m,n;

    while(scanf("%d%d",&m,&n)!=EOF){

        memset(dp_row,0,sizeof(dp_row));

        memset(dp_col,0,sizeof(dp_col));

        int tmp;

        for(int i=1;i<=m*n;i++){

            scanf("%d",&tmp);

            if(i%n==1){

                dp_row[i][0]=0;

                dp_row[i][1]=tmp;

            }else{

                dp_row[i][0]=max(dp_row[i-1][0],dp_row[i-1][1]);

                dp_row[i][1]=dp_row[i-1][0]+tmp;

            }

        }

        for(int i=1;i<=m;i++){

            dp_col[i][0]=max(dp_col[i-1][0],dp_col[i-1][1]);

            dp_col[i][1]=dp_col[i-1][0]+max(dp_row[i*n][0],dp_row[i*n][1]);

        }

        printf("%d\n",max(dp_col[m][0],dp_col[m][1]));

    }

    return 0;

}