[hihocoder]01背包 accept 与 time limit exceed的代码比较

 1 #include <iostream>
 2 #include <algorithm>
 3 using namespace std;
 4 
 5 int dp[501][100001];
 6 int need[501];
 7 int value[501];
 8 
 9 int main()
10 {
11     int num,amount;
12     cin>>num>>amount;
13     for(int i=1;i<=num;i++)
14     {
15         cin>>need[i]>>value[i];
16     }
17 
18     for(int j=1;j<=num;j++)
19     {
20         for(int i=0;i<=amount;i++)
21         {
22             if(i<need[j])
23             {
24                 dp[j][i]=dp[j-1][i];
25             }
26             else
27             {
28                 dp[j][i]=max(dp[j-1][i],(dp[j-1][i-need[j]]+value[j]));
29             }
30         }
31     }
32 
33     cout<<dp[num][amount]<<endl;
34     return 0;
35 }

 1 #include <iostream>
 2 #include <algorithm>
 3 using namespace std;
 4 
 5 int dp[100001][501];
 6 int need[501];
 7 int value[501];
 8 
 9 int main()
10 {
11     int num,amount;
12     cin>>num>>amount;
13     for(int i=1;i<=num;i++)
14     {
15         cin>>need[i]>>value[i];
16     }
17 
18     for(int j=1;j<=num;j++)
19     {
20         for(int i=0;i<=amount;i++)
21         {
22             if(need[j]>i)
23             {
24                 dp[i][j]=dp[i][j-1];
25             }
26             else
27             {
28                 dp[i][j]=max(dp[i][j-1],(dp[i-need[j]][j-1]+value[j]));
29             }
30         }
31     }
32 
33     cout<<dp[amount][num]<<endl;
34     return 0;
35 }