NYOJ 题目234 吃土豆

<span style="font-family: Tahoma, Arial, sans-serif, simsun; widows: auto; background-color: rgb(255, 255, 255);">描述</span>

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

分析：题意看不懂的看图，图给的很清楚（如取81，则他左右的两个数是不能取的，他的上一行下一行也是不能取的），可以先求出每一行的最大值，map+=max（map[i][j-2],map[i][j-3]），然后再把每一行的最大值进行计算找到最合适的那个，dp[i]=max(dp[i-2],dp[i-3])+max(map[i][m+1],map[i][m+2]);。

AC代码：

<span style="font-size:18px;">#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int dp[510];
int map [510][510];
int main()
{
int n,m,i,j;
while(~scanf("%d%d",&n,&m))
{
memset(map,0,sizeof(map));
memset(dp,0,sizeof(dp));
for(i=3; i<n+3; i++)
for(j=3; j<m+3; j++)
{
scanf("%d",&map[i][j]);
map[i][j]+=max(map[i][j-2],map[i][j-3]);
}
for(i=3; i<n+3; i++)
{
dp[i]=max(dp[i-2],dp[i-3])+max(map[i][m+1],map[i][m+2]);
}
printf("%d\n",dp[n+2]);
}

return 0;
}</span>