hdu 1171 Big Event in HDU (01背包，母函数)

最新推荐文章于 2020-09-04 01:51:25 发布

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原文链接：http://www.cnblogs.com/GetcharZp/p/9536563.html

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本文探讨了一个关于设施在计算机学院和软件学院之间的公平分配问题，通过01背包问题的算法解决，确保两个学院获得的设施价值尽可能相等，且计算机学院的价值不低于软件学院。

Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 51181 Accepted Submission(s): 17486

Problem Description
Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don't know that Computer College had ever been split into Computer College and Software College in 2002.
The splitting is absolutely a big event in HDU! At the same time, it is a trouble thing too. All facilities must go halves. First, all facilities are assessed, and two facilities are thought to be same if they have the same value. It is assumed that there is N (0<N<1000) kinds of facilities (different value, different kinds).

Input
Input contains multiple test cases. Each test case starts with a number N (0 < N <= 50 -- the total number of different facilities). The next N lines contain an integer V (0<V<=50 --value of facility) and an integer M (0<M<=100 --corresponding number of the facilities) each. You can assume that all V are different.
A test case starting with a negative integer terminates input and this test case is not to be processed.

Output
For each case, print one line containing two integers A and B which denote the value of Computer College and Software College will get respectively. A and B should be as equal as possible. At the same time, you should guarantee that A is not less than B.

Sample Input
2
10 1
20 1
3
10 1
20 2
30 1
-1

Sample Output
20 10
40 40

C/C++（01背包）：

 1 #include <map>
 2 #include <queue>
 3 #include <cmath>
 4 #include <vector>
 5 #include <string>
 6 #include <cstdio>
 7 #include <cstring>
 8 #include <climits>
 9 #include <iostream>
10 #include <algorithm>
11 #define INF 0x3f3f3f3f
12 using namespace std;
13 const int MAX = 1e6 + 10;
14 
15 int n, sum, dp[MAX], v, num, val[MAX], cnt;
16 
17 int main()
18 {
19     while (scanf("%d", &n), n > 0)
20     {
21         sum = cnt = 0;
22         memset(dp, 0, sizeof(dp));
23         while (n --)
24         {
25             scanf("%d%d", &v, &num);
26             while (num --)
27             {
28                 val[cnt ++] = v;
29                 sum += v;
30             }
31         }
32         for (int i = 0; i < cnt; ++ i)
33             for (int j = (sum >> 1); j >= val[i]; -- j)
34                 dp[j] = max(dp[j], dp[j - val[i]] + val[i]);
35         printf("%d %d\n", sum - dp[sum >> 1], dp[sum >> 1]);
36     }
37     return 0;
38 }

C/C++（母函数）：

 1 #include <map>
 2 #include <queue>
 3 #include <cmath>
 4 #include <vector>
 5 #include <string>
 6 #include <cstdio>
 7 #include <cstring>
 8 #include <climits>
 9 #include <iostream>
10 #include <algorithm>
11 #define INF 0x3f3f3f3f
12 using namespace std;
13 const int MAX = 1e6 + 10;
14 
15 int n, last2, last, a[MAX], b[MAX], num[1010], val[1010], ans;
16 
17 int main()
18 {
19     while (scanf("%d", &n), n > 0)
20     {
21         for (int i = 0; i < n; ++ i) scanf("%d%d", &val[i], &num[i]);
22         a[0] = 1, last = 0;
23         for (int i = 0; i < n; ++ i)
24         {
25             last2 = last + val[i] * num[i];
26             memset(b, 0, sizeof(int) * (last2 + 1));
27             for (int j = 0; j <= num[i]; ++ j)
28                 for (int k = 0; k <= last; ++ k)
29                     b[k + j * val[i]] += a[k];
30             memcpy(a, b, sizeof(int) * (last2 + 1));
31             last = last2;
32         }
33         for (ans = (last>>1); ans >= 0 && a[ans] == 0; -- ans);
34         printf("%d %d\n", last - ans, ans);
35     }
36     return 0;
37 }