nyoj 吃土豆 234 （双层DP）

吃土豆

时间限制：1000 ms  |           内存限制：65535 KB

难度：4

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

4 611 0 7 5 13 978 4 81 6 22 41 40 9 34 16 1011 22 0 33 39 6

样例输出

242

//题意：//给你n行m列的矩阵，矩阵上的数代表此处有多少豆子，问最多能吃多少豆子//但吃豆子有一定的规则。//规则：：如果你吃了某处的豆子，那么你不能再吃它左边和右边的两处的豆子，并且不能吃它上下两行的豆子//思路：//双层DP，先求出每行的最大值，再求全部的最大值。

#include<stdio.h>#include<string.h>int max(int x,int y){return x>y?x:y;}int dp[510];int a[510][510];int main(){int n,m,i,j,w;while(scanf("%d%d",&n,&m)!=EOF){memset(dp,0,sizeof(dp));memset(a,0,sizeof(a));for(i=3;i<n+3;i++){for(j=3;j<m+3;j++){scanf("%d",&w);a[i][j]=max(a[i][j-2],a[i][j-3])+w;}}for(i=3;i<n+3;i++)dp[i]=max(dp[i-2],dp[i-3])+max(a[i][m+1],a[i][m+2]);printf("%d\n",max(dp[n+1],dp[n+2]));}return 0;}