hdu1171 01背包

Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36600    Accepted Submission(s): 12715

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

 

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 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

 

题意：给你n件物品，让你分成尽量价值相等的两堆

思路：01背包，输入的时候记录一个totalvalue，然后你只需要计算出容量为（totalvalue/2）的背包能装的最大价值，就是其中一堆的价值，然后用totalvalue - dp[ half_totalvalue ]就是另一堆的价值。

贴一下代码：

#include
#include
#include
#include
using namespace std;

int value[5050];
int dp[255555];
int number[5050];

int main(){

    int n;
    while(scanf("%d",&n) && n >= 0){
        memset(dp,0,sizeof(dp));
        int totalvalue = 0;
        for(int i = 1; i <= n; i++){
            scanf("%d%d",&value[i],&number[i]);
            totalvalue += value[i]*number[i];
        }

        int half = totalvalue >> 1;
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= number[i]; j++)
                for(int k = half; k >= value[i]; k--)
                    dp[k] = max(dp[k],dp[k-value[i]]+value[i]);
        cout << totalvalue - dp[half] << " " << dp[half] << endl;

    }
}