zoukankan      html  css  js  c++  java
  • HDU-1171 Big Event in HDU (多重背包)

    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
     

     
    思路:
    多重背包,把总价值的一半做背包体积即可。
     

     
    Code:
     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 const int maxn = 1000 + 10;
     4 int val[maxn], num[maxn], dp[maxn*50*100];
     5 
     6 int main() {
     7     int n;
     8     while(~scanf("%d", &n), n >= 0) {
     9         memset(dp, 0, sizeof(dp));
    10         int sum = 0;
    11         for (int i = 0; i < n; ++i) {
    12             scanf("%d%d", &val[i], &num[i]);
    13             sum += val[i] * num[i];
    14         }
    15         int allsum = sum;
    16         sum /= 2;
    17         for (int i = 0; i < n; ++i) {
    18             if (val[i] * num[i] >= sum)
    19                 for (int j = val[i]; j <= sum; ++j)
    20                     dp[j] = max(dp[j], dp[j-val[i]] + val[i]); 
    21             else {
    22                 int m = val[i] * num[i];
    23                 for (int k = 1; k * val[i] <= m; k <<= 1) {
    24                     int v = k * val[i];
    25                     for (int j = sum; j >= v; --j)
    26                         dp[j] = max(dp[j], dp[j-v] + v);
    27                     m -= v;
    28                 }
    29                 for (int j = sum; j >= m; --j)
    30                     dp[j] = max(dp[j], dp[j-m] + m);
    31             }
    32         }
    33         int ans = dp[sum];
    34         printf("%d %d
    ", allsum - ans, ans);
    35     }
    36 
    37     return 0;
    38 }
  • 相关阅读:
    x32dbg之AttachHelper插件
    x32dbg插件之APIInfo
    x32dbg之Scylla脱壳插件
    x32dbg插件之strongOD(又名SharpOD)
    x32dbg新型插件之loli(萝莉)
    7 个超棒的监控工具
    成为程序员前需要做的10件事
    改良程序的11个技巧
    旧衣物捐献地址和注意事项
    一件衣服好不好,看看标签就知道
  • 原文地址:https://www.cnblogs.com/robin1998/p/6426967.html
Copyright © 2011-2022 走看看