DP:Cow Bowling（POJ 3176）

题目很简单的，我就不翻译了，简单来说就是储存每一行的总数，类似于状态压缩

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #define MAX(a,b) a>b?a:b
 4 
 5 int Cow_Map[351][351];
 6 
 7 void Search(const int);
 8 
 9 int main(void)
10 {
11     int N, i, line;
12     while (~scanf("%d", &N))
13     {
14         Cow_Map[0][0] = 0;
15         for (line = 1; line <= N; line++)//读图
16             for (i = 0; i < line; i++)
17                 scanf("%d", &Cow_Map[line][i]);
18         Search(N);
19     }
20     return 0;
21 }
22 
23 void Search(const int N)
24 {
25     int line, i, ans = 0;
26     for (line = 2; line <= N; line++)
27     {
28         Cow_Map[line][0] = Cow_Map[line][0] + Cow_Map[line - 1][0];
29         for (i = 1; i < line - 1; i++)
30             Cow_Map[line][i] = (MAX(Cow_Map[line - 1][i - 1], Cow_Map[line - 1][i])) + Cow_Map[line][i];
31         Cow_Map[line][line - 1] = Cow_Map[line][line - 1] + Cow_Map[line - 1][line - 2];
32     }
33     for (i = 0; i < N; i++)
34         ans = MAX(Cow_Map[N][i], ans);
35     printf("%d
", ans);
36 }