洛谷 P3399 丝绸之路

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:

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

using namespace std;

int n,m,a[1001],b[1001],f[1001][1001],ans = 1999999999;

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]);
    for(int i = 1;i <= m; i++)
        scanf("%d",&b[i]);
    f[1][1] = b[1] * a[1];
    for(int i = 1;i <= m; i++) f[1][i] = min(f[1][i-1],a[1] * b[i]);
    for(int i = 2;i <= n; i++)
        for(int j = i;j <= m; j++)
                f[i][j] = min(f[i][j-1],f[i-1][j-1] + a[i] * b[j]);
    for(int i = n;i <= m; i++)
        ans = min(f[n][i],ans);
    printf("%d",ans);
    return 0;
}