HDU 3905 Sleeping ---dp

来源：互联网发布：sechk是什么软件编辑：程序博客网时间：2024/05/16 17:29

题目地址： http://acm.hdu.edu.cn/showproblem.php?pid=3905

AC代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <string>#include <vector>#include <list>#include <deque>#include <queue>#include <iterator>#include <stack>#include <map>#include <set>#include <algorithm>#include <cctype>using namespace std;typedef long long LL;const int N=1002;const LL II=1000000007;const int INF=0x3f3f3f3f;const double PI=3.141592653;int f[N][N],g[N][N];int val[N],sum[N];int n,m,l,ans;int main(){int i,j,k;while(~scanf("%d%d%d",&n,&m,&l)){    sum[0]=0;for(i=1;i<=n;i++){scanf("%d",&val[i]);sum[i]=sum[i-1]+val[i];}k=n-m;if(k<=0){puts("0");continue;}memset(f,0,sizeof(f));memset(g,0,sizeof(g));ans=0;for(i=l;i<=n;i++){for(j=l;j<=k&&j<=i;j++){f[i][j]=g[i-l][j-l]+sum[i]-sum[i-l];if(f[i-1][j-1]){f[i][j]=max(f[i-1][j-1]+val[i],f[i][j]);}g[i][j]=max(g[i-1][j],f[i][j]);ans=max(ans,g[i][j]);}}printf("%d\n",ans);}return 0;}/*f[i][j]表示前i分钟学习j分钟并且第i分钟是学习的。g[i][j]表示第i分钟可以学也可以睡。转移方程f[i][j]=mmax(f[i-1][j-1]+val[i],g[i-l][j-l]+sum[i]-sum[i-l];]);g[i][j]=mmax(g[i-1][j],f[i][j]);*/