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。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <climits>
using namespace std;
int d[510][510],dp[510][2];
int mx[510];
int main() {
    int r,c,i,j,temp,ans;
    while(~scanf("%d %d",&r,&c)) {
        for(i = 1; i <= r; i++) {
            memset(dp,0,sizeof(dp));
            mx[i] = INT_MIN;
            for(j = 1; j <= c; j++) {
                scanf("%d",d[i]+j);
                dp[j][0] = max(dp[j-1][0],dp[j-1][1]);
                dp[j][1] = dp[j-1][0] + d[i][j];
                temp = max(dp[j][0],dp[j][1]);
                mx[i] = max(temp,mx[i]);
            }
        }
        memset(dp,0,sizeof(dp));
        ans = INT_MIN;
        for(i = 1; i <= r; i++) {
            dp[i][0] = max(dp[i-1][0],dp[i-1][1]);
            dp[i][1] = dp[i-1][0] + mx[i];
            temp = max(dp[i][0],dp[i][1]);
            if(temp > ans) ans = temp;
        }
        printf("%d
",ans);
    }
    return 0;
}