HDU2845_Beans【不连续的最大子段和】【元素压缩】

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Beans

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2964 Accepted Submission(s): 1427

Problem Description
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 ?

Input
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<=200000.

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

Sample Input
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

Sample Output
242

Source

2009 Multi-University Training Contest 4 - Host by HDU

题目大意：给你一个矩阵，不能选择每行中相邻的数字，也不能选当前行的上一

行和下一行，问使所选数和最大的值是多少？

思路：用元素压缩的思想。先把2维矩阵降为1维数组。对每行求出不相邻的数字

最大和是多少，把几个数字和缩成一个数。再对所有行求出不相邻的行数字最大

和是多少。

对于每行求出不相邻的数字最大和的状态转移方程为

dp[i+1] = max{吃i达到的最大值，不吃i达到的最大值+第i+1个数}

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;const int MAXN = 200000;int dpa[MAXN+20],dpb[MAXN+20],row[MAXN+20];int main(){    int M,N,num;    while(~scanf("%d%d",&M,&N))    {        memset(row,0,sizeof(row));        for(int i = 0; i < M; i++)        {            dpa[0] = dpb[0] = 0;            for(int j = 0; j < N; j++)            {                scanf("%d",&num);                dpa[j+1] = max(dpa[j],dpb[j]);// dp[j+1] 是到j为止，不吃j所能吃到的最大值                dpb[j+1] = dpa[j] + num;//吃j所能吃到的最大值            }            row[i] = max(dpa[N],dpb[N]);        }        dpa[0] = dpb[0] = 0;        for(int i = 0; i < M; i++)        {            dpa[i+1] = max(dpa[i],dpb[i]);            dpb[i+1] = dpa[i] + row[i];        }        int ans = max(dpa[M],dpb[M]);        printf("%d\n",ans);    }    return 0;}

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