Big Event in HDU

Time Limit : 10000/5000ms (Java/Other)???Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 84???Accepted Submission(s) : 38

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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 ??題目的意思就是把這些設(shè)備盡量的平分成兩份，先輸出大的，再輸出小的。
? ? ?總價(jià)值（所有v[i]*m[i]之和）的一半作為背包的容量，把每一種中的每一個(gè)設(shè)備都看成一塊石頭。按01背包的解法，打表更新。使答案非常接近總質(zhì)量的一半（就是在不超過總質(zhì)量一半的前提下，dp【k】越大就是越接近）。
? ? ?外面的兩個(gè)循環(huán)就是遍歷每一個(gè)設(shè)備，相當(dāng)于01背包的外循環(huán)：for（i=1;i<=石頭的數(shù)量;i++）。最里面的循環(huán)讓容量從sum到v【i】（可以不到0，因?yàn)?到v【i】這部分，放不了任何石頭，不更新，也就不需要判斷v【i】是否大于k了） #include <iostream> #include <algorithm> #include <string> #include <cstring> using namespace std; int main() { int T; int v[55], m[55]; while (cin >> T && T >= 0) { int i; int s = 0; for (i = 1; i <= T; i++) { cin >> v[i] >> m[i]; s = s + v[i] * m[i]; } int sum = s / 2; int dp[250005]; memset(dp, 0, sizeof(dp)); int j; for (i = 1; i <= T; i++)//對(duì)每一種遍歷 { for (j = 1; j <= m[i]; j++)//有幾個(gè)就遍歷幾個(gè) {//把它變成01背包，總共有T*m[i]個(gè)石頭，遍歷 int k; for (k = sum; k >= v[i]; k--)//最小的石頭就是v[i]大，只要到v[i]就可以了 { dp[k] = max(dp[k], dp[k - v[i]] + v[i]); } } } if (dp[sum] > s - dp[sum]) cout << dp[sum] << " " << s - dp[sum] << endl; else cout << s - dp[sum] << " " << dp[sum] << endl; } return 0; }

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轉(zhuǎn)載于:https://www.cnblogs.com/caiyishuai/p/8325880.html

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