HDU 5445 Food Problem（多重背包+二进制优化）

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 using namespace std;
 5 const int INF = 0x3f3f3f3f;
 6 
 7 int n,m,p,tot;
 8 int val[2005],cost[2005];
 9 int dp[50005];
10 
11 int solve1()
12 {
13     for(int i=0;i<=p+100;i++)  dp[i] = INF;
14     dp[0] = 0;
15     for(int i=0;i<tot;i++)
16         for(int j=p+100;j>=val[i];j--)
17         dp[j] = min(dp[j],dp[j-val[i]]+cost[i]);
18     int ans = INF;
19     for(int i=p;i<=p+100;i++)
20         ans = min(ans, dp[i]);
21     return ans;
22 }
23 
24 
25 int solve2(int v)
26 {
27     memset(dp,0,sizeof dp);
28     for(int i=0;i<tot;i++)
29         for(int j=50000;j>=cost[i];j--)
30         dp[j] = max(dp[j],dp[j-cost[i]]+val[i]);
31     for(int i=1;i<=50000;i++)
32         if(dp[i]>=v) return i;
33     return INF;
34 }
35 
36 int main()
37 {
38     //freopen("in.txt","r",stdin);
39     int T;
40     scanf("%d",&T);
41     while(T--)
42     {
43         scanf("%d%d%d",&n,&m,&p);
44         tot = 0;
45         for(int i=1;i<=n;i++)
46         {
47             int t,u,v;
48             scanf("%d%d%d",&t,&u,&v);
49             for(int k=1;v;k<<=1)
50             {
51                 int num = min(k,v);
52                 val[tot] = num*t;
53                 cost[tot++] = num*u;
54                 v -= num;
55             }
56         }
57         int v = solve1();
58         tot = 0;
59         for(int i=1;i<=m;i++)
60         {
61             int x,y,z;
62             scanf("%d%d%d",&x,&y,&z);
63             for(int k=1;z;k<<=1)
64             {
65                 int num = min(k,z);
66                 cost[tot] = num*y;
67                 val[tot++] = num*x;
68                 z -= num;
69             }
70         }
71         if(v==INF) {puts("TAT");continue;}
72         int ans = solve2(v);
73         if(ans==INF)  puts("TAT");
74         else printf("%d\n",ans);
75     }
76     return 0;
77 }