Nyoj Eating potatoes (dynamic planning)

Describe
Bean-eating is a interesting game, everyone owns an m*n matrix, which are filled with different qualities beans. Meantime, there is the only one bean in any 1*1 grid. Now your want to eat the beans and collect the qualities, but everyone must obey by the following rules:if you eat the BEA n 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 do you eat and then get?
Input
There is a few cases. In each case, there is the 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 beans isn ' t beyond, and 1<=m,n<=500.
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

If it is a line, that is, when the n==1, Dp[i] said from the beginning to the number of I can obtain the maximum and, there is Dp[i]=max (Dp[i-1], dp[i-2]+a[i]); This is the case of each row, calculate the maximum sum of each line can be obtained, and then vertical, the adjacent row can not be taken, you can use the same method to find the largest sum of the whole matrix, the overall idea is to first calculate each line and then two dimensions into one dimension.

AC Code:

# include <cstdio># include <cstring># include <algorithm>using namespace Std;int a[510][510], dp_y[ 510][510], Ans[510];int main () {int n, m, I, J, K;while (scanf ("%d%d", &n, &m)!=eof) {memset (A, 0, sizeof (a)); for (i= 1; i<=n; i++) {for (j=1; j<=m; J + +) {scanf ("%d", &a[i][j]);}} memset (dp_y, 0, sizeof (dp_y)), for (I=1; i<=n; i++) {dp_y[i][1]=a[i][1];for (j=2; j<=m; J + +) {Dp_y[i][j]=max (dp_y[i ][j-1], dp_y[i][j-2]+a[i][j]);}} memset (ans, 0, sizeof (ans)), ans[1]=dp_y[1][m];for (i=2; i<=n; i++) {Ans[i]=max (ans[i-1], ans[i-2]+dp_y[i][m]);} printf ("%d\n", Ans[n]);} return 0;}

Nyoj Eating potatoes (dynamic planning)