题目传送门

解题思路:

f[i][j]表示第j天到第i个城市消耗的最小体力值,对于每种状态,我们可能从两种状态转移过来:
1.我们可以选择前一天没动,即f[i][j-1]
2.可以选择前一天动了,即f[i-1][j-1]

还有,第1座城市比较特殊,所以要单独处理一下.

AC代码:

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 
 5 using namespace std;
 6 
 7 int n,m,a[1001],b[1001],f[1001][1001],ans = 1999999999;
 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     for(int i = 1;i <= m; i++)
20         scanf("%d",&b[i]);
21     f[1][1] = b[1] * a[1];
22     for(int i = 1;i <= m; i++) f[1][i] = min(f[1][i-1],a[1] * b[i]);
23     for(int i = 2;i <= n; i++)
24         for(int j = i;j <= m; j++)
25                 f[i][j] = min(f[i][j-1],f[i-1][j-1] + a[i] * b[j]);
26     for(int i = n;i <= m; i++)
27         ans = min(f[n][i],ans);
28     printf("%d",ans);
29     return 0;
30 }