HDU

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Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43705    Accepted Submission(s): 15030

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

210 120 1310 1 20 230 1-1

 

Sample Output

20 1040 40

 

Author

lcy

 

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题目大意：

给出物品的价值V和数量M，将所有物品分成A,B两份，要求尽量平均且A不小于B。

思路：

分成两堆，其中较小的那一堆的最大值为总价值的一半。所以将背包容量为价值总和的一半，转化成01背包问题。

附上AC代码：

#include<bits/stdc++.h>using namespace std;const int maxn = 5000 + 5;int cnt[maxn];int dp[250000];int N, V, M;int cur, sum;int main() {    ios :: sync_with_stdio(false);    while(cin >> N && N > 0) {        cur = sum = 0;        memset(dp, 0, sizeof(dp) );        memset(cnt, 0, sizeof(cnt) );        for(int i = 0; i < N; i++) {            cin >> V >> M;            for(int j = 0; j < M; j++) {                cnt[cur++] = V;                sum += V;            }        }        for(int i = 0; i< cur; i++) {            for(int j = sum / 2; j >= cnt[i]; j--) {                dp[j] = max(dp[j], dp[j - cnt[i]] + cnt[i]);              }        }        cout<< sum - dp[sum / 2] << ' ' << dp[sum / 2] << endl;    }    return 0;}