UVa1366 Martian Mining
首先应该想到一点, 如果一段管道被中途截断了, 那么这段管道就是毫无意义的, 没了它答案也不会变差.
那么, 决策就可以看成每次修一条管道(从下到上或者从右到左), 状态可以定为每次考虑一个矩阵(顺序从小到大)
其次,很明显这个题满足最优子结构性质.
那么就不难写出方程了
1 #include <cstdio>
2 #include <cstring>
3 #include <iostream>
4 #include <algorithm>
5 using namespace std;
6 const int MAXN = 5e2 + 20;
7 inline int read()
8 {
9 int x = 0; char ch = getchar();
10 while(!isdigit(ch)) ch = getchar();
11 while(isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
12 return x;
13 }
14 int N, M;
15 int squ[MAXN][MAXN][2];
16 int f[MAXN][MAXN];
17
18 int main()
19 {
20 while(cin>>N>>M, N)
21 {
22 memset(squ, 0, sizeof(squ));
23 for(int i = 1; i <= N; i++)
24 for(int j = 1; j <= M; j++)
25 squ[i][j][0] = squ[i][j - 1][0] + read();
26
27 for(int i = 1; i <= N; i++)
28 for(int j = 1; j <= M; j++)
29 squ[i][j][1] = squ[i - 1][j][1] + read();
30
31 memset(f, 0, sizeof(f));
32 for(int i = 1; i <= N; i++)
33 for(int j = 1; j <= M; j++)
34 f[i][j] = max(f[i][j], max(f[i][j - 1] + squ[i][j][1], f[i - 1][j] + squ[i][j][0]));
35 cout<<f[N][M]<<endl;
36 }
37 return 0;
38 }
