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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 ?
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
Source
2009 Multi-University Training Contest 4 - Host by HDU
思路:注意状态转移方程,还有就是行的转移方程和列的相似,还有注意的是数组开大点
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define N 200005 int dpx[N],dpy[N]; int main() { int n,m,i,j,s; while(~scanf("%d%d",&n,&m)) { memset(dpx,0,sizeof(dpx)); memset(dpy,0,sizeof(dpy)); for(i=2;i<=n+1;i++) { //memset(dpy,0,sizeof(dpy)); //这里为什么可以注释掉呢,因为会被覆盖 for(j=2;j<=m+1;j++) { scanf("%d",&s); dpy[j]=max(dpy[j-1],dpy[j-2]+s); //dpy表示这一行从左到右能取到的最大的数和 } dpx[i]=max(dpx[i-1],dpx[i-2]+dpy[1+m]); //dpx表示从以上行中能取到数的最大的和 } printf("%d\n",dpx[n+1]); } return 0; }
HDU 2845 Beans(dp)
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