NYOJ 234 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 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

来源

2009 Multi-University Training Contest 4

上传者

张洁烽

　　

解题：先求每一行的最大值，然后把这些最大值再求一次最大值。这道题目还是有点意思啦。慢慢想想还是可以解出来的。想再快点的话可以再开个数组，把后面那个循环里面的内容放到读入循环里面去。

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <climits>

 5 using namespace std;

 6 int d[510][510],dp[510][2];

 7 int mx[510];

 8 int main() {

 9     int r,c,i,j,temp,ans;

10     while(~scanf("%d %d",&r,&c)) {

11         for(i = 1; i <= r; i++) {

12             memset(dp,0,sizeof(dp));

13             mx[i] = INT_MIN;

14             for(j = 1; j <= c; j++) {

15                 scanf("%d",d[i]+j);

16                 dp[j][0] = max(dp[j-1][0],dp[j-1][1]);

17                 dp[j][1] = dp[j-1][0] + d[i][j];

18                 temp = max(dp[j][0],dp[j][1]);

19                 mx[i] = max(temp,mx[i]);

20             }

21         }

22         memset(dp,0,sizeof(dp));

23         ans = INT_MIN;

24         for(i = 1; i <= r; i++) {

25             dp[i][0] = max(dp[i-1][0],dp[i-1][1]);

26             dp[i][1] = dp[i-1][0] + mx[i];

27             temp = max(dp[i][0],dp[i][1]);

28             if(temp > ans) ans = temp;

29         }

30         printf("%d\n",ans);

31     }

32     return 0;

33 }

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