http://acm.hdu.edu.cn/showproblem.php?pid=2845&&最大不连续数和
2012-03-26 09:37
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Beans
[b]Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1176 Accepted Submission(s): 615
[/b]
[align=left]Problem Description[/align]
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 ?
[align=left]Input[/align]
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.
[align=left]Output[/align]
For each case, you just output the MAX qualities you can eat and then get.
[align=left]Sample Input[/align]
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
[align=left]Sample Output[/align]
242题意:最大不连续和问题,如果选中(x,y),那么x-1和x+1行不能选,(x,y-1)和(x,y+1)也不能选,,,状态转移方程b[i]=max(b[i-2]+b[i],b[i-1])AC代码:
#include<iostream> #include<string.h> #include<string> #include<algorithm> #define N 200005 using namespace std; int a ,b ; int main() { int n,m; while(cin>>n>>m) { memset(b,0,sizeof(b)); for(int i=1;i<=n;++i) { for(int j=1;j<=m;++j) cin>>b[j]; for(int j=2;j<=m;++j) b[j]=max(b[j-2]+b[j],b[j-1]); a[i]=b[m]; } for(int i=2;i<=n;++i) a[i]=max(a[i-2]+a[i],a[i-1]); cout<<a <<endl; }return 0; }
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