NYOJ 题目17 单调递增最长子序列（DP） hdu 题目2845 Bean

吃土豆

时间限制：1000 ms | 内存限制：65535 KB

难度：4

描述

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 ?

输入

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<=500.

输出

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

样例输入

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

样例输出

二次dp，

1.首先单独对每行的数据进行DP处理，得到一个最大值；

2,.每行的最大值又组成一个新的数组，再次dp求最大值

 
#define N 505
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int MAX(int x,int y){
	return x>y?x:y;
}
int main()
{
	int m,n,i,j,k,max,a[N][N],dp[N][2],f[N]; 
	
	while(scanf("%d%d",&m,&n)!=EOF){
		
		for(i=0;i<m;i++) 
			for(j=0;j<n;j++)
				scanf("%d",&a[i][j]);
	
		for(i=0;i<m;i++) {
			dp[0][0]=0; dp[0][1]=a[i][0];		
			for(j=1;j<n;j++){
				dp[j][0] = MAX(dp[j-1][0],dp[j-1][1]);
				dp[j][1] = dp[j-1][0] + a[i][j];
			}		
			f[i] = MAX(dp[n-1][0],dp[n-1][1]);
		}
		dp[0][0] = 0;  dp[0][1] = f[0];
		
		for(j=1;j<m;j++){		
			dp[j][0] = MAX(dp[j-1][0],dp[j-1][1]);
			dp[j][1] = dp[j-1][0] + f[j];
		}				
		printf("%d\n",MAX(dp[m-1][0],dp[m-1][1]));
	}
	
	return 0;
}

HDU 题目

Beans

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2144 Accepted Submission(s): 1081

Problem Description

Now, how much qualities can you eat and then get ?

Input

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

代码要严谨，直接上面的代码通不过；

改进代码：

#define N 200005
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int MAX(int x,int y){
    return x>y?x:y;
}
int a[N],dp[N][2],f[N]; 
int main()
{
    int m,n,i,j,k,max;        
    while(scanf("%d%d",&m,&n)!=EOF){    
        for(i=0;i<m;i++) {
            for(j=0;j<n;j++)
                scanf("%d",&a[j]);                            
            dp[0][0]=0; dp[0][1]=a[0];    
            for(j=1;j<n;j++){
                dp[j][0] = MAX(dp[j-1][0],dp[j-1][1]);
                dp[j][1] = dp[j-1][0] + a[j];
            }        
            f[i] = MAX(dp[n-1][0],dp[n-1][1]);                                                        
        }
        dp[0][0] = 0;  dp[0][1] = f[0];
        for(j=1;j<m;j++){        
            dp[j][0] = MAX(dp[j-1][0],dp[j-1][1]);
            dp[j][1] = dp[j-1][0] + f[j];
        }    
        printf("%d\n",MAX(dp[m-1][0],dp[m-1][1]));
    }
    
    return 0;
}