NYOJ-234-DP(吃土豆)

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

 

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int main(){int a[550][550],dp[550];int i,j,n,m,map;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",&map);a[i][j]=max(a[i][j-2],a[i][j-3])+map;}}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;}

 

还是这个好，比上面的占用内存少多了。。。值得学习

 

 #include<stdio.h>int max(int m,int *a){int i,f[550];for(i=1;i<=m;i++){if(i==1)  f[i]=a[i];else if(i==2)  f[i]=a[i]>arr[i-1]?a[i]:a[i-1];else {f[i]=(f[i-2]+a[i])>f[i-1]?f[i-2]+a[i]:f[i-1];}}return f[m];}//非递归推算法/*int max(int m,int *a){if(m==1)return a[1];if(m==2)return a[2]=a[1]>a[2]?a[1]:a[2];return max(m-2,a)+a[m]>max(m-1,a)?max(m-2,a)+a[m]:max(m-1,a);}*///递归分制算法int main(){int a[510],b[510],i,j,m,n;while(scanf("%d %d",&n,&m)==2){for(i=1;i<=n;i++){for(j=1;j<=m;j++){scanf("%d",&a[j]);}b[i]=max(m,a);}printf("%d\n",max(n,b)); }return 0;}