HDU 2845 Beans DP

最新推荐文章于 2024-05-25 16:47:04 发布

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最新推荐文章于 2024-05-25 16:47:04 发布

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分类专栏： ACM-DP

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7月22-8月21多校联合训练期间，会根据实际负载关闭部分模块，若有不便，请谅解~

Beans

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2585 Accepted Submission(s): 1274

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

Source

2009 Multi-University Training Contest 4 - Host by HDU

状态转移dp[i]=max(dp[i-1],dp[i-2]+dp[i]);

横竖求一下就好

然后取每一行的最大得b[i] ,又可看作一行相同处理

#define DeBUG
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <string>
#include <set>
#include <sstream>
#include <map>
#include <list>
#include <bitset>
using namespace std ;
#define zero {0}
#define INF 0x3f3f3f3f
#define EPS 1e-6
#define TRUE true
#define FALSE false
typedef long long LL;
const double PI = acos(-1.0);
//#pragma comment(linker, "/STACK:102400000,102400000")
inline int sgn(double x)
{
    return fabs(x) < EPS ? 0 : (x < 0 ? -1 : 1);
}
#define N 200005
int n, m;
int a[N];
int dp[N];
int dpl[N];
int b[N];
int main()
{
#ifdef DeBUGs
    freopen("C:\\Users\\Sky\\Desktop\\1.in", "r", stdin);
#endif
    while (scanf("%d%d", &n, &m) + 1)
    {
        memset(a, 0, sizeof(a));
        memset(dp, 0, sizeof(dp));
        for (int i = 0; i < n; i++)
        {
            dp[i] = dpl[i] = -INF;
        }
        for (int k = 0; k < n; k++)
        {
            for (int j = 0; j < m; j++)
            {
                scanf("%d", &a[j]);
            }
            dp[0] = a[0];
            dp[1] = max(a[0],a[1]);
            for (int i = 2; i < m; i++)
            {
                dp[i] = max(dp[i - 1], dp[i - 2] + a[i]);
            }
            b[k] = dp[m - 1];
        }
        dpl[0] = b[0];
        dpl[1]=max(b[0],b[1]);
        for (int i = 2; i < n; i++)
        {
            dpl[i] = max(dpl[i - 1], dpl[i - 2] + b[i]);
        }
        printf("%d\n", dpl[n - 1]);
    }

    return 0;
}