传送门

f[i] 表示送前 i 头牛过去再回来的最短时间

f[i] = min(f[i], f[j] + sum[i j] + m) (0 <= j < i)

——代码

1 #include <cstdio> 2 #include <iostream> 3 4 const int MAXN = 2501, INF = 12345678; 5 int n, m; 6 int sum[MAXN], f[MAXN]; 7 8 inline long long read() 9 16 17 inline int min(int x, int y) 18 21 22 int main() 23 33 for(i = 1; i <= n; i++) 34 38 printf("%d\n", f[n] m); 39 return 0; 40 }
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