HDU 2845(动态规划-最大非连续区间)

最新推荐文章于 2021-08-17 21:53:34 发布

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本文介绍了一款名为“豆子游戏”的问题，通过动态规划算法解决如何在遵循特定规则的情况下从矩阵中获取最大豆子价值的问题。文章详细阐述了解决方案的思路，并提供了完整的代码实现。

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问题描述:

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 ?

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

题目题意:题目给了我们一个矩阵，每格有一个值，我们要在题目要求情况下（满足俩个要求)，求出最总和。

取了(x,y) 则(x,y-1) (x,y+1)不能取，x-1行和x+1行不能取。

题目分析:我们先处理去每一行可以得到的最大值（求最大非连续区间），然后保存起来，按照行的规则，又是一个最大非连续区间。

dp[i]=max(dp[i-1]+a[i],dp[i-2]);

代码如下:

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;

const int maxn=2*1e5+100;
int clomn[maxn],row[maxn],dp[maxn];

int main()
{
    int n,m;
    while (scanf("%d%d",&n,&m)!=EOF) {
          memset (dp,0,sizeof (dp));
          for (int i=1;i<=n;i++) {
              for (int j=1;j<=m;j++) {
                  scanf("%d",&clomn[j]);
              }
              dp[1]=clomn[1];
              for (int j=2;j<=m;j++) {
                 dp[j]=max(dp[j-2]+clomn[j],dp[j-1]);
              }
              row[i]=dp[m];
          }
          dp[1]=row[1];
          for (int i=2;i<=n;i++)
            dp[i]=max(dp[i-1],dp[i-2]+row[i]);
          printf("%d\n",dp[n]);
    }
    return 0;
}