hdu2182Frog(动态规划)

Problem Description
A little frog named Fog is on his way home. The path's length is N (1 <= N <= 100), and there are many insects along the way. Suppose the
original coordinate of Fog is 0. Fog can stay still or jump forward T units, A <= T <= B. Fog will eat up all the insects wherever he stays, but he will
get tired after K jumps and can not jump any more. The number of insects (always less than 10000) in each position of the path is given.
How many insects can Fog eat at most?
Note that Fog can only jump within the range [0, N), and whenever he jumps, his coordinate increases.

Input
The input consists of several test cases.
The first line contains an integer T indicating the number of test cases.
For each test case:
The first line contains four integers N, A, B(1 <= A <= B <= N), K (K >= 1).
The next line contains N integers, describing the number of insects in each position of the path.

Output
each test case:
Output one line containing an integer - the maximal number of insects that Fog can eat.

Sample Input
1
4 1 2 2
1 2 3 4

Sample Output
8

无脑的dpAC代码：

include

include

using namespace std;
int main()
{
int t;
cin>>t;
while(t--)
{
int N,A,B,K;
cin>>N>>A>>B>>K;
int a[120]={0};
for(int i=0;i<N;i++)
cin>>a[i];
int dp[120][120]; //dp[i][k]表示在第i+1个位置的第k次跳后的最大值
memset(dp,-1,sizeof(dp));
dp[0][0]=a[0];
for(int i=0;i<N;i++)
{
for(int j=A;j<=B;j++)
{
for(int k=0;k<=max(K,101);k++)
{
if(dp[i][k]!=-1&&i+j<N) dp[i+j][k+1]=max(dp[i][k]+a[i+j],dp[i+j][k+1]);
else continue;
}
}
}
int max1=0;
for(int i=0;i<N;i++)
{
for(int j=1;j<=K;j++)
{
if(dp[i][j]>max1) max1=dp[i][j];
}
}
cout<<max1<<endl;
}
return 0;
}

但凡使用dp的，稍微有点难度的基本都会用到max()