[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 = 2501, INF = 12345678;
int n, m;
int sum[MAXN], f[MAXN];

inline long long read()
{
    long long x = 0, f = 1;
    char ch = getchar();
    for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
    for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
    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[0] = m;
    for(i = 1; i <= n; i++)
    {
        x = read();
        sum[i] = x + sum[i - 1];
    }
    for(i = 1; i <= n; i++)
    {
        f[i] = INF;
        for(j = 0; j < i; j++) f[i] = min(f[i], f[j] + sum[i - j] + m);
    }
    printf("%d
", f[n] - m);
    return 0;
}