描述
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 ?
输入
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<=500.
输出
For each case, you just output the MAX qualities you can eat and then get.
样例输入
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
样例输出
242
如果是一行的话，就是说当n==1时，dp[i]表示从开始到第i个数所能取得的最大和，则有dp[i]=max(dp[i-1], dp[i-2]+a[i]);这是每一行的情况，算出每一行的所能得到的最大和，接下来再竖着来，相邻的行不能取，就能用同样的方法求出整个矩阵的最大和，总体思路就是先算每一行再将二维化成一维。
AC代码：
# 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;
}