HDU 2845 Beans

最新推荐文章于 2019-03-28 22:43:44 发布

原创最新推荐文章于 2019-03-28 22:43:44 发布 · 354 阅读

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#ACM新手上路 #HDU #c++ #dp #入门题

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本文介绍了一种豆子游戏的算法实现，玩家在一个M*N的矩阵中吃豆子，每颗豆子有不同的质量。根据特定的游戏规则，文章详细阐述了如何通过动态规划的方法来计算能够吃到的最大豆子质量。

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Beans

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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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

题意

吃豆子，每个豆子有不同的质量，假如吃了(x,y)位置的豆子，那么就不能吃x-1行和x+1行的豆子，也不能吃(x,y-1)和(x,y+1)位置的豆子。求能吃到的豆子的最大质量。

分析
先找到每一行可以吃到的最大质量，再由每行的最大质量推全部的最大质量。
状态转移方程：
行最大
linedp[i] = max(linedp[j-1],linedp[j-2] + num[i][j])(i为行,j为列)
全体最大
dp[i] = max(dp[i-1],dp[i-2] + linedp[i])

AC代码如下

#include <cstdio>
#include <cstring>
#define maxn 200005
#define max(a,b) a>b?a:b
using namespace std;

int a[maxn],dp[maxn],ldp[maxn];
int m,n;

int main()
{
    while(~scanf("%d%d",&m,&n))
    {
        for(int i = 1 ; i <= m ; i++)
        {
            dp[0] = 0;
            for(int j = 1 ; j <= n; j++)
                scanf("%d",&a[j]);
            dp[1] = a[1];
            for(int j = 2 ; j <= n ; j++)
                dp[j] = max(dp[j-1],dp[j-2] + a[j]);
            ldp[i] = dp[n];
        }
        dp[1] = ldp[1];
        for(int i = 2 ; i <= m ; i++)
            dp[i] = max(dp[i-1],dp[i-2] + ldp[i]);
        printf("%d\n",dp[m]);

    }
    return 0;
}