[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

 1 #include <cstdio>
 2 #include <iostream>
 3 
 4 const int MAXN = 2501, INF = 12345678;
 5 int n, m;
 6 int sum[MAXN], f[MAXN];
 7 
 8 inline long long read()
 9 {
10     long long x = 0, f = 1;
11     char ch = getchar();
12     for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
13     for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
14     return x * f;
15 }
16 
17 inline int min(int x, int y)
18 {
19     return x < y ? x : y;
20 }
21 
22 int main()
23 {
24     int i, j, x;
25     n = read();
26     m = read();
27     sum[0] = m;
28     for(i = 1; i <= n; i++)
29     {
30         x = read();
31         sum[i] = x + sum[i - 1];
32     }
33     for(i = 1; i <= n; i++)
34     {
35         f[i] = INF;
36         for(j = 0; j < i; j++) f[i] = min(f[i], f[j] + sum[i - j] + m);
37     }
38     printf("%d\n", f[n] - m);
39     return 0;
40 }
View Code

 

posted @ 2017-05-26 10:45  zht467  阅读(97)  评论(0)  编辑  收藏  举报