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:

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 
 5 using namespace std;
 6 
 7 int n,m,a[2501],sum[2501],f[2501];
 8 
 9 inline int min(int s,int d) {
10     if(s < d) return s;
11     return d;
12 }
13 
14 int main() {
15     memset(f,0x3f3f3f,sizeof(f));
16     scanf("%d%d",&n,&m);
17     for(int i = 1;i <= n; i++) {
18         scanf("%d",&a[i]);
19         sum[i] = sum[i-1] + a[i];
20     }
21     f[0] = 0;
22     for(int i = 1;i <= n; i++) {
23         f[i] = sum[i];
24         for(int j = 1;j < i; j++)
25             f[i] = min(f[i],f[j] + sum[i-j] + m);
26         f[i] += m;
27     }
28     printf("%d",f[n]);
29     return 0;
30 }