This solution is based on 0/1 knapsack problem.
dp[i][j][k] is the maximum amount of songs with i songs while k unit time is left in disk j.
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#include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <string> using namespace std; int N,T,M; int song[800]; int dp[800][100][100]; void DP() { for(int i=0;i<N;i++) { for(int j=0;j<=M;j++) { for(int k=0;k<=T;k++) { if(k<song[i]) dp[i+1][j][k]=dp[i][j][k]; else if(k>=song[i]) dp[i+1][j][k]=max(dp[i][j][k-song[i]]+1,dp[i][j][k]); if(j) dp[i+1][j][k]=max(dp[i][j-1][T-song[i]]+1,dp[i+1][j][k]); } } } } int main() { int tt; cin>>tt; while(tt--) { cin>>N>>T>>M; cin>>song[0]; for(int i=1;i<N;i++) { scanf(", %d",&song[i]); } memset(dp,0,sizeof(dp)); DP(); int res = 0; for(int i=0;i<=N;i++) { res = max(res,dp[i][M][0]); } cout<<res<<"\n"; if(tt) cout<<"\n"; } return 0; } |