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
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
#include
#include
#include
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; }
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