題意

　　給定 $n \times m$ 的網格圖, 每個格子有 $w$ 個財寶.

　　每次從左上角出發到右下角, 將途中經過的所有格子中的財寶至多拿一個.

　　問最少多少次能拿完所有財寶.

　　$n, m \le 1000$ .

分析

　　根據 Dilworth 定理, 最少鏈劃分 = 最大反鏈長度.

　　從左下角到右上角進行 DP .

實現

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cstdlib>
 4 #include <cctype>
 5 #include <algorithm>
 6
 
 using namespace std;
 7 #define F(i, a, b) for (register int i = (a); i <= (b); i++)
 8 #define P(i, a, b) for (register int i = (a); i >= (b); i--)
 9 #define LL long long
10 inline int rd(void) {
11     int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -1;
12
 
     int x = 0; for (; isdigit(c); c = getchar()) x = x*10+c-‘0‘; return x*f;
13 }
14 
15 const int N = 1005;
16 
17 int n, m, w[N][N];
18 LL f[N][N];
19 
20 int main(void) {
21     #ifndef ONLINE_JUDGE
22         freopen("bzoj3997.in", "r", stdin);
23     #endif
24     
25     for (int nT = rd(); nT > 0
 
; nT--) {
26         n = rd(), m = rd();
27         F(i, 1, n) F(j, 1, m) w[i][j] = rd();
28     
29         memset(f, 0, sizeof f);
30         P(i, n, 1)
31             F(j, 1, m) {
32                 f[i][j] = max(f[i][j-1], f[i+1][j]);
33                 f[i][j] = max(f[i][j], f[i+1][j-1] + w[i][j]);
34             }
35         printf("%lld\n", f[1][m]);
36     }
37     
38     return 0;
39 }

[BZOJ 3997] 組合數學