洛谷 P2904 [USACO08MAR]跨河River Crossing

题目传送门

解题思路:

f[i]表示把i只羊渡河的最少时间,则本次可能带1~i只羊过河,然后转移方程就出来了.

AC代码:

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

int n,m,a[2501],sum[2501],f[2501];

inline int min(int s,int d) {
    if(s < d) return s;
    return d;
}

int main() {
    memset(f,0x3f3f3f,sizeof(f));
    scanf("%d%d",&n,&m);
    for(int i = 1;i <= n; i++) {
        scanf("%d",&a[i]);
        sum[i] = sum[i-1] + a[i];
    }
    f[0] = 0;
    for(int i = 1;i <= n; i++) {
        f[i] = sum[i];
        for(int j = 1;j < i; j++)
            f[i] = min(f[i],f[j] + sum[i-j] + m);
        f[i] += m;
    }
    printf("%d",f[n]);
    return 0;
}