HDU-2845-Beans（简单DP）

最新推荐文章于 2019-01-28 00:54:01 发布

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本文介绍了一款名为Beans的游戏中的动态规划算法实现。游戏规则是在一个M*N的矩阵中吃豆子并累计分数，但吃掉某豆子后其周围特定位置的豆子将不能被吃。文章详细阐述了如何通过DP_X[]和DP_Y[]两个数组进行状态转移，最终求得最大可获得的分数。

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Beans

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4488 Accepted Submission(s): 2115

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.
这里写图片描述

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

题意：吃掉（i，j）后，他的左右两格和上下两行都不能再吃了，求获取的最大数值

坑点 1<=M*N<=200000，这个范围也是相当可以的

思路：DP_X[]取一行的最大值，DP_Y[]取每行最大值中的最大值。

状态转移方程:
DP_X[0]=DP_X[1]=DP_Y[0]=DP_Y[1]=0//注意(0,0),(1,1)都初始化为0,从(2,2)开始接收数据
DP_X[i]=max(DP_X[j-2]+num,DP_X[j-1]);
DP_Y[i]=max(DP_Y[i-2]+DP_X[M+1],DP_Y[i-1]

代码

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<vector>
using namespace std;
//简单DP
const int maxn=200005;
int DP_X[maxn];
int DP_Y[maxn];
int main()
{
    int N,M;
    while(~scanf("%d%d",&N,&M))
    {
        DP_X[0]=DP_X[1]=0;
        DP_Y[0]=DP_Y[1]=0;
        for(int i=2; i<=N+1; i++)
        {
            for(int j=2; j<=M+1; j++)
            {
                int flag;
                scanf("%d",&flag);
                DP_X[j]=max(DP_X[j-2]+flag,DP_X[j-1]);//应该不用比较自身吧
            }
            DP_Y[i]=max(DP_Y[i-2]+DP_X[M+1],DP_Y[i-1]);
        }
        printf("%d\n",DP_Y[N+1]);
    }
    return 0;
}