HDU-1171 Big Event in HDU (多重背包)

Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 44674    Accepted Submission(s): 15369

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

#include <bits/stdc++.h>using namespace std;int a[51], c[51], dp[130000];int main(){int n;while(scanf("%d", &n) != EOF){if(n < 0) return 0;int tot = 0;for(int i = 1; i <= n; ++i){scanf("%d %d", &a[i], &c[i]);tot += a[i] * c[i];}int ans = tot / 2;memset(dp, 0, sizeof(dp));for(int i = 1; i <= n; ++i){for(int j = 1; j <= c[i]; j *= 2){int v = j * a[i];for(int k = ans; k >= v; --k){dp[k] = max(dp[k], dp[k - v] + v);}c[i] -= j;}if(c[i]){int v = c[i] * a[i];for(int k = ans; k >= v; --k){dp[k] = max(dp[k], dp[k - v] + v);}}}printf("%d %d\n", tot - dp[ans], dp[ans]);}}/*题意：50种设备，每种设备不超过100个，每种设备的价值不超过50，将这些设备分成价值尽量接近的两部分。思路：转换成多重背包问题。背包容量为所有设备总价值的一半sum/2。那么一部分就是dp[sum/2],一部分是sum-dp[sum/2]。坑点：最后是文件尾是负数，不是-1.WA到三观毁灭。*/