HDU 1171 Big Event in HDU dp背包

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本文介绍了一个关于设施价值在两个学院间进行尽可能等价分配的问题。通过01背包问题的算法解决，确保计算机学院和软件学院获得的设施价值尽可能相等且前者不低于后者。给出了解决该问题的具体代码实现。

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40976 Accepted Submission(s): 14090

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

利用整除的性质，满足A is not less than B ,然后扩展背包数目，开始01背包
坑点就是看清题目中的dp数，数组要开大，否则会超时

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
int dp[300000];
int w[6000];

int main() {
    int n;
    while(scanf("%d",&n)==1&&n>=0) {
        memset(dp,0,sizeof(dp));
       // memset(w,0,sizeof(w));
        int a,b;
        int t=1;
        int sum=0;
        for(int i=1;i<=n;i++) {
            scanf("%d%d",&a,&b);
            sum+=a*b;
            while(b--) {
                w[t++]=a;
                //sum+=a;
            }
        }
        int m=sum/2;
        for(int i=1;i<t;i++) {
            for(int j=m;j>=w[i];j--) {
                dp[j]=max(dp[j],dp[j-w[i]]+w[i]);
            }
        }
        printf("%d %d\n",sum-dp[m],dp[m]);

    }

    return 0;
}