母函数练习2

最新推荐文章于 2022-02-21 16:53:54 发布

原创最新推荐文章于 2022-02-21 16:53:54 发布 · 302 阅读

0 ·

CC 4.0 BY-SA版权

数学专栏收录该内容

7 篇文章

订阅专栏

题目：hdu1171

题意：有n个物品，每个物品有它的重量和它的数目。求怎么尽可能把它们分成两部分。

解答：写出其母函数，然后模拟（也可以用完全背包）

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
const int MAXN = 250005;
int c1[MAXN],c2[MAXN];
struct Event
{
    int V,M;
};
Event event[MAXN];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        if(n<0)
            break;
        int sum = 0;
        for(int i = 1;i <= n;i++)
        {
             scanf("%d%d",&event[i].V,&event[i].M);
             sum += event[i].V * event[i].M;
        }
        memset(c1,0,sizeof(c1));
        for(int i = 0;i <= event[1].M;i++)
            c1[i*event[1].V] = 1;
        int m = event[1].V * event[1].M;
        memset(c2,0,sizeof(c2));
        for(int i = 2;i <= n;i++)
        {
            for(int j = 0 ;j <= MAXN;j++)
            for(int k = 0;k <= event[i].M;k++)
                c2[k*event[i].V+j] += c1[j];
            for(int i = 0;i <= MAXN;i++)
            {
                c1[i] = c2[i];
                c2[i] = 0;
            }
        }
        int t = sum;
        sum = sum/2;
        int a,b;
        for(int i = sum;i>=0;i--)
            if(c1[i])
        {
            a = i;
            break;
        }
        for(int i = sum;i<=250000;i++)
            if(c1[i])
        {
            b = i;
            break;
        }
         int ans;
         double tmp = (double)t / 2;
         if(tmp - a < b - tmp)
            ans = a;
        else
            ans = b;
        int temp = max(ans,t - ans);
        printf("%d %d\n",temp,t - temp);
    }
    return 0;
}