Beans (最大不连续子序列和)
发布时间:2020-12-15 01:57:31 所属栏目:Java 来源:网络整理
导读:Beans Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2637????Accepted Submission(s): 1302 Problem Description Bean-eating is an interesting game,everyone owns an M*N matrix,which is
BeansTime Limit: 2000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
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 ?
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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.
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Output
For each case,you just output the MAX qualities you can eat and then get.
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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
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Sample Output
242
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Source
2009 Multi-University Training Contest 4 - Host by HDU
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Recommend
gaojie
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这道题是说取一个数字,那么相邻的两行的所有数字和同一行的相邻两列的数字就不能再取了。
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在同一行里面,相邻的数字不能同时取,
那么每一行都有会有一个最大值,那么可以把每一行最大值看成一个数字,再求一次,相当于在一行里面求最大不连续子序列的和,只是这里的每一个数字是每一行的最大值。先求出每一行的最大值然后我们再看我们到底应该去取哪一行。
dp[i]=max(dp[i-2]+a[i],dp[i-1])
dp[i]代表到i时的最大子序列的和,对于每一个数字,我只有取或者不取两种状态,如果取,那么我的最大值是从dp[i-2]再加上本身的数值,如果不取,那么我最大值是dp[i-1]。
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感觉这个和01背包很像,取a[i]这个位置和不取a[i]这个位置
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1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <algorithm> 5 6 #define LL long long 7 using namespace std; 8 const int MAXN = 200005; 9 10 int n,m; 11 int dp[MAXN]; 12 int a[MAXN]; 13 14 15 int main() 16 { 17 // freopen("../in.txt","r",stdin); 18 while (~scanf("%d%d",&n,&m)) 19 { 20 memset(a,0,sizeof(a)); 21 for (int i=1;i<=n;i++) 22 { 23 for (int j=1;j<=m;j++) 24 { 25 scanf("%d",&a[j]); 26 } 27 for (int j=2;j<=m;j++) 28 { 29 a[j] = max(a[j-2]+a[j],a[j-1]); 30 } 31 dp[i] = a[m]; 32 } 33 for (int i=1;i<=n;i++) 34 { 35 dp[i] = max(dp[i-2]+dp[i],dp[i-1]); 36 } 37 printf("%dn",dp[n]); 38 } 39 return 0; 40 } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |
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