USACO 2008 Mar Silver 3.River Crossing 动态规划水题

Code：

#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
const int maxn = 2500 + 4;
const int inf = 100000000;
int f[maxn], sumv[maxn];
int main()
{
    freopen("r.in","r",stdin);
    freopen("r.out","w",stdout);
    int n, m;
    scanf("%d%d",&n,&m);
    for(int i = 1;i <= n; ++i)scanf("%d",&sumv[i]), sumv[i] += sumv[i - 1];
    for(int i = 1;i <= n; ++i)
    {
        f[i] = inf;
        for(int j = 0;j < i; ++j) f[i] = min(f[i], f[j] + sumv[i - j]);
        f[i] += 2 * m;
    }
    printf("%d",f[n] - m);
    fclose(stdin);
    fclose(stdout);
    return 0;
}