NY–234 — 吃馬鈴薯 [二維動態規劃]

 

吃馬鈴薯 時間限制：1000 ms  |  記憶體限制：65535 KB難度：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 ?

 

 

輸入

 

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.

輸出

For each case, you just output the MAX qualities you can eat and then get.

範例輸入

 

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

範例輸出

 242

 

Code:

 

吃掉一個馬鈴薯之後,左右的兩個和上下一行的都不能吃..

這一題可以看做是兩層的dp, 是求最長遞增子序列的升級版

在限制條件下, 先在每一行求出最大的和,

再把這些和看做是一行,再DP一次即可

代碼是在輸入的時候就進行和的儲存, 取i-1的值和i-1與i的和的最大值存在i裡面

這樣最後得到的一定是這一行的最大和

 

#include<stdio.h>#include<string.h>#define max(a,b) a>b?a:b int val[200005],dp[200005];int main(){int r,c,i,j;while(scanf("%d%d",&r,&c)!=EOF){memset(val,0,sizeof(val));memset(dp,0,sizeof(dp));for(i=0;i<r;i++){for(j=0;j<c;j++){scanf("%d",&val[j]);if(j==1) val[1] = max(val[1],val[0]);if(j>1)  val[j] = max(val[j-1],val[j]+val[j-2]);}dp[i] = val[c-1];}dp[1] = max(dp[0],dp[1]);for(i=2;i<r;i++)dp[i] = max(dp[i-1],dp[i]+dp[i-2]);printf("%d\n",dp[r-1]);}return 0;}