状态机模型解题
根据题意出发来分析这个问题:有两种状态(来这家(设状态为1)/不来这家(设状态为0))
所以:
#include <iostream>
using namespace std;
const int N = 1e5 + 10;
int arr[N];
int f[N][2];
main(void)
{
int t;
cin >> t;
while (t --)
{
int n;
cin >> n;
for (int i = 1;i <= n; i ++)
cin >> arr[i];
for (int i = 1; i <= n; i ++)
{
f[i][0] = max(f[i - 1][1], f[i - 1][0]);
f[i][1] = f[i - 1][0] + arr[i];
}
cout << max(f[n][0], f[n][1]) << endl;
}
return 0;
}
线性dp 也是可以做的
#include <iostream>
using namespace std;
const int N = 1e5 + 10;
int arr[N];
int f[N][2];
main(void)
{
int t;
cin >> t;
while (t --)
{
int n;
cin >> n;
for (int i = 1;i <= n; i ++)
cin >> arr[i];
f[1] = arr[1];
for (int i = 2; i <= n; i ++)
f[i] = max(f[i - 2] + arr[i], f[i - 1]);
cout << f[n] << endl;
}
return 0;
}