HDU-1864&&HDU-2602（01背包问题）

 1 #include <iostream>
 2 #include <queue>
 3 #include <cstdio>
 4 #include <algorithm>
 5 #include <cmath>
 6 #include <cstring>
 7 #include <queue>
 8 #include <map>
 9 #include <vector>
10 #define INF 0x3f3f3f3f
11 #define FRE() freopen("in.txt","r",stdin)
12 
13 using namespace std;
14 typedef long long ll;
15 const int maxn = 500;
16 double dp[maxn];
17 double money,a[maxn];
18 
19 bool judge(char op)
20 {
21     if(op==‘A‘||op==‘B‘||op==‘C‘)
22         return true;
23     return false;
24 }
25 
26 
27 void init(int n,int& index)
28 {
29     char op,ch;
30     while(n--)
31     {
32         int x,ok = 1;
33         double tmp,sumA = 0,sumB = 0,sumC = 0;
34         cin>>x;
35         for(int i = 0; i<x; i++)
36         {
37             cin>>ch>>op>>tmp;
38             if(judge(ch))
39             {
40                 if(ch==‘A‘)sumA += tmp;
41                 if(ch==‘B‘)sumB += tmp;
42                 if(ch==‘C‘)sumC += tmp;
43             }
44             else
45                 ok = 0;
46         }
47         if(ok && (sumA<=600.0 && sumB<=600.0 && sumC<=600.0 && (sumA+sumB+sumC) <=1000.0))
48             a[index++] = (sumA+sumB+sumC)*100.0;
49     }
50     return;
51 }
52 
53 
54 int main()
55 {
56     int n,index = 0,in;
57     while(cin>>money>>n)
58     {
59         if(n==0)break;
60         index = 0;
61         memset(a,0,sizeof(a));
62         init(n,index);
63 //        for(int i = 0; i<index; i++)
64 //        {
65 //            cout<<a[i]<<" ";
66 //        }
67 //        cout<<endl<<endl;
68         memset(dp,0,sizeof(dp));
69         in = 1;
70         for(int i = 0; i<index; i++)//枚举每一个可以报销的票
71         {
72             for(int j = 0; j<=i; j++)
73             {
74                 if(dp[j]+a[i] <= money*100.0)
75                 {
76                     dp[i] = max(dp[i],dp[j]+a[i]);//每一个票的位置上对应一个dp维护该位置上的最大报销额度
77                 }
78             }
79         }
80         double ans = 0;
81         for(int i = 0; i<index; i++)
82         {
83             ans = max(ans,dp[i]);
84         }
85         printf("%.2f\n",ans/100.0);
86     }
87     return 0;
88 }

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 
 5 using namespace std;
 6 const int maxn = 1005;
 7 int N,V;
 8 int wet[maxn],val[maxn];
 9 int dp[maxn];
10 
11 int main()
12 {
13     //freopen("in.txt","r",stdin);
14     int T;
15     scanf("%d",&T);
16     while(T--)
17     {
18         memset(wet, 0, sizeof(wet));
19         memset(val, 0, sizeof(val));
20         memset(dp, 0, sizeof(dp));
21         scanf("%d%d",&N,&V);
22         for(int i = 0; i < N; i++)
23             scanf("%d",&val[i]);
24         for(int i = 0; i < N; i++)
25             scanf("%d",&wet[i]);
26         for(int i = 0; i < N; i++)//枚举每一块骨头
27         {
28             for(int j = V; j >= 0; j--)//枚举背包体积区间的每一个大小
29             {
30                 if(j >= wet[i])//如果背包体积大于骨头的体积
31                     dp[j] = max(dp[j], dp[j - wet[i]] + val[i]);//更新该体积所能装下的最大的价值
32             }
33         }
34         printf("%d\n",dp[V]);
35     }
36     return 0;
37 }