HDU 2845 Beans （DP）

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

4 611 0 7 5 13 978 4 81 6 22 41 40 9 34 16 1011 22 0 33 39 6

 

Sample Output

242

题目可以看成是一个二维的，每一维的解法都是一个DP的过程，也就是一个数组，取第i个数就不能取与他相邻的数，求和最大。

可以设两个数组d[i],f[i]为别表示第i个数取或者不取时的和最大。

d[i] = f[i-1]+a[i] , f[i] = max(f[i-1],d[i-1]) 。f[1] = 0, d[1] = a[1] .

#include <stdio.h>#include <string.h>#include <algorithm>#include <math.h>using namespace std;typedef long long LL;const int MAX=0x3f3f3f3f;const int maxn = 200005;int n, m;int d[maxn], f[maxn], a[maxn], b[maxn];int DP(int *c, int len) {    d[1] = c[1], f[1] = 0;    for(int i = 1; i <= len; i++) {        d[i] = f[i-1] + c[i];        f[i] = max(f[i-1], d[i-1]);    }    return max(f[len], d[len]);}int main(){    while(~scanf("%d%d", &n, &m)) {        for(int i = 1; i <= n; i++) {            for(int j = 1; j <= m; j++)                scanf("%d", &a[j]);            b[i] = DP(a, m);        }        printf("%d\n", DP(b, n));    }    return 0;}