[Usaco07Jan] Running
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题目链接
题解:dp
#include <cstdio>#include <cstring>#include <cmath>#include <string>#include <iostream>#include <algorithm>using namespace std;const int N=10005, M=505;int f[N][M],n,m,d[N];//第i分钟疲劳为j的跑步距离int main() { cin>>n>>m; for(int i=1;i<=n;i++) cin>>d[i]; for(int i=1;i<=n;i++){ f[i][0]=max(f[i][0],f[i-1][0]);//疲劳值为0时休息 for(int j=1;j<=n;j++) { f[i][j]=f[i-1][j-1]+d[i];//跑 if(i-j>0) f[i][0]=max(f[i][0],f[i-j][j]);//休息,因为疲劳值要恢复0,所以从f[i-j][j]转移过来 } } cout<<f[n][0]<<endl; return 0;}
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