[luoguP2904] [USACO08MAR]跨河River Crossing（DP）

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

 #include <cstdio>
 #include <iostream>

 const int MAXN = , INF = ;
 int n, m;
 int sum[MAXN], f[MAXN];

 inline long long read()
 {
     long long x = , f = ;
     char ch = getchar();
     for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -;
     for(; isdigit(ch); ch = getchar()) x = (x << ) + (x << ) + ch - '';
     return x * f;
 }

 inline int min(int x, int y)
 {
     return x < y ? x : y;
 }

 int main()
 {
     int i, j, x;
     n = read();
     m = read();
     sum[] = m;
     for(i = ; i <= n; i++)
     {
         x = read();
         sum[i] = x + sum[i - ];
     }
     for(i = ; i <= n; i++)
     {
         f[i] = INF;
         for(j = ; j < i; j++) f[i] = min(f[i], f[j] + sum[i - j] + m);
     }
     printf("%d\n", f[n] - m);
     return ;
 }