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HDU 2845 Beans (DP)
Beans
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64uDescription
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 ?
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
4 611 0 7 5 13 978 4 81 6 22 41 40 9 34 16 1011 22 0 33 39 6
Sample Output
242
思路：
STEP_1:先以行为单位来算，设DP数组存的是以此位置为起点能得到的最大值，在此前提下，DP[i][j]因为相邻的不能走,所以要么加上DP[i][j + 2]，要么加上DP[i][j + 3] (假设不越界)，取最大的那个就好。从右往左依次计算这一行每个元素的DP值，记录下最大的，放到DP[i][0]里。
STEP_2:重复第一步，只不过对象换成了DP数组里每行0号单元的值，因为此单元放的是此行内的最大值。从上往下，每次选定一行后设此行为最终答案里最下面那行，因为选定一行之后它的上面和下面那一行就废掉了，所以每一行要么加上它上面第二行，要么加上它上面第三行。同理，记录下最大值，最大值即答案。(实际上答案不是最后一行就是倒数第二行，因为没有负数，只会越加越大，不过只有1行的时候就会越界，虽然HDU上的数据貌似没1行的)。
1 #include<stdio.h> 2 #include<stdlib.h> 3 #include<string.h> 4 #define MAX 200105 5 6 int main(void) 7 { 8 int n,m; 9 int max,temp_1,temp_2,ans;10 11 while(scanf("%d%d",&n,&m) != EOF)12 {13 int dp[n + 10][m + 10];14 15 memset(dp,0,sizeof(dp));16 for(int i = 1;i <= n;i ++)17 for(int j = 1;j <= m;j ++)18 scanf("%d",&dp[i][j]);19 20 for(int i = 1;i <= n;i ++)21 {22 max = dp[i][m];23 for(int j = m;j >= 1;j --)24 {25 dp[i][j] += dp[i][j + 2] > dp[i][j + 3] ? dp[i][j + 2] : dp[i][j + 3];26 max = max > dp[i][j] ? max : dp[i][j];27 }28 dp[i][0] = max;29 }30 31 max = dp[1][0];32 dp[3][0] += dp[1][0];33 for(int i = 4;i <= n;i ++)34 {35 dp[i][0] += dp[i - 2][0] > dp[i - 3][0] ? dp[i - 2][0] : dp[i - 3][0];36 max = max > dp[i][0] ? max : dp[i][0];37 }38 39 printf("%d\n",max);40 }41 42 return 0;43 }
HDU 2845 Beans (DP)
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