吃土豆
2016-04-08 10:16
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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
样例输出
242
考虑对于某行某列元素,row[i][j]表示加上位置为i,j的土豆的质量的i行j列最大的和
列的最大值:row[i][j]=max(row[i][j-2]+row[i][j-3])+val
看图说话:
假设红色的格子为i行j列,那么它的前面有两种选择方案:
1、选择蓝色格子
2、选择黄色格子
那么该行最大的和是什么呢?
由于n列、n-1列具有状态无关性(n-1列的状态影响不了n列的状态),很显然等于max(row[i]
,row[i][n-1])
同理对于dp[i] (i行的最大值)
dp[i]=max(dp[i-2],dp[i-3])+max_row[i]
看图说话:
max土豆质量=max(dp[m],dp[m-1])
为了方便计算,我的代码把n,m扩大了2
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
考虑对于某行某列元素,row[i][j]表示加上位置为i,j的土豆的质量的i行j列最大的和
列的最大值:row[i][j]=max(row[i][j-2]+row[i][j-3])+val
看图说话:
假设红色的格子为i行j列,那么它的前面有两种选择方案:
1、选择蓝色格子
2、选择黄色格子
那么该行最大的和是什么呢?
由于n列、n-1列具有状态无关性(n-1列的状态影响不了n列的状态),很显然等于max(row[i]
,row[i][n-1])
同理对于dp[i] (i行的最大值)
dp[i]=max(dp[i-2],dp[i-3])+max_row[i]
看图说话:
max土豆质量=max(dp[m],dp[m-1])
为了方便计算,我的代码把n,m扩大了2
#include<iostream> #include<cstdio> #include<cstring> using namespace std; int e[505][505],dp[505]; int n,m; int main(){ while(scanf("%d%d",&m,&n)!=EOF){ memset(e,0,sizeof(dp)); memset(dp,0,sizeof(dp)); int i,j,map; for(i=3;i<m+3;++i){ for(j=3;j<n+3;++j){ scanf("%d",&map); e[i][j]=max(e[i][j-2],e[i][j-3])+map; } } for(i=3;i<m+3;++i){ dp[i]=max(dp[i-2],dp[i-3])+max(e[i][n+1],e[i][n+2]); } printf("%d\n",max(dp[m+1],dp[m+2])); } return 0; }
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