Hdu 5445 Food Problem (2015长春网络赛 ACM/ICPC Asia Regional Changchun Online)

题目链接:

  Hdu  5445 Food Problem 

题目描述:

  有n种甜点,每种都有三个属性(能量,空间,数目),有m辆卡车,每种都有是三个属性(空间,花费,数目)。问至少运输p能量的甜点,花费最小是多少?

解题思路:

  明显可以看出是多重背包搞两次,但是数据范围太大了,背包要到2*1e6,感觉会TLe。还是呆呆的写了一发,果断超啊!然后滚回去看背包九讲课件了,看到了二进制压缩的时候,感觉可以搞这个题目。试了一下果然AC,原本物品数目是100*100,二进制压缩以后也就是100*log2100个左右,然后进行01背包就OK咯!

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <algorithm>
 5 using namespace std;
 6 const int INF = 0x3f3f3f3f;
 7 const int maxn = 55000;
 8 const int N = 2000;
 9 int dp[maxn], x, y, c1, c2;
10 int dcos[N], dval[N], tcos[N], tval[N];
11 void display ()
12 {
13     for (int i=0; i<c1+200; i++)
14         dp[i] = INF;
15     dp[0] = 0;
16     for (int i=0; i<x; i++)
17         for (int j=c1+200; j>=dval[i]; j--)
18             dp[j] = min (dp[j], dp[j-dval[i]]+dcos[i]);
19     c2 = dp[c1];
20     for (int i=c1; i<c1+200; i++)
21         c2 = min (dp[i], c2);
22 }
23 void solve ()
24 {
25     memset (dp, 0, sizeof(dp));
26     for (int i=0; i<y; i++)
27         for (int j=50000; j>=tcos[i]; j--)
28             dp[j] = max (dp[j], dp[j-tcos[i]]+tval[i]);
29     for (int i=1; i<=50000; i++)
30         if (dp[i] >= c2)
31         {
32             printf ("%d\n", i);
33             return ;
34         }
35     printf ("TAT\n");
36 }
37 int main ()
38 {
39     int t, n, m;
40     scanf ("%d", &t);
41     while (t --)
42     {
43         scanf ("%d %d %d", &n, &m, &c1);
44         int a, b, c;
45         x = y = 0;
46         for (int i=0; i<n; i++)
47         {
48             scanf ("%d %d %d", &a, &b, &c);
49             for (int k=1; c; k*=2)
50             {
51                 int num = min (k, c);
52                 dcos[x] = num * b;
53                 dval[x++] = num * a;
54                 c -= num;
55             }
56         }
57         for (int i=0; i<m; i++)
58         {
59             scanf ("%d %d %d", &a, &b, &c);
60             for (int k=1; c; k*=2)
61             {
62                 int num = min (k, c);
63                 tcos[y] = num * b;
64                 tval[y++] = num * a;
65                 c -= num;
66             }
67         }
68         display ();
69         solve ();
70     }
71     return 0;
72 }

 

posted @ 2015-09-16 17:23  罗茜  阅读(364)  评论(0编辑  收藏  举报