NYOJ234吃土豆（双层动态规划）

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

思路：首先找出每一行能吃的最大土豆数，在按照列dp一次就可以找出最大数了

#include<stdio.h>#include<string.h> #define max(x,y)(x>y?x:y)int map[510][510];int dp[510][510][2];int maxR[510];int dp2[510][2];int main(){int m,n,i,j,k;while(scanf("%d%d",&m,&n)!=EOF){memset(dp,0,sizeof(dp));memset(dp2,0,sizeof(dp2));for(i=1;i<=m;i++)for(j=1;j<=n;j++)scanf("%d",&map[i][j]);for(k=1;k<=m;k++){for(i=1;i<=n;i++){dp[k][i][0]=max(dp[k][i-1][0],dp[k][i-1][1]);dp[k][i][1]=dp[k][i-1][0]+map[k][i];}}for(i=1;i<=m;i++)maxR[i]=max(dp[i][n][0],dp[i][n][1]);for(i=1;i<=m;i++){dp2[i][0]=max(dp2[i-1][0],dp2[i-1][1]);dp2[i][1]=dp2[i-1][0]+maxR[i];}int res=max(dp2[m][0],dp2[m][1]);printf("%d\n",res);}return 0;}