poj 2576 简单二维背包

最新推荐文章于 2022-11-28 19:03:53 发布

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本文介绍了一个经典的体重分配问题——Tug of War。该问题要求将n个人分成两组进行拔河比赛，使得两组的人数差不超过1且总重量尽可能接近。通过使用动态规划的方法，文章提供了一种有效的解决方案，并附带了完整的C语言代码实现。

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题目要求：

n个人，每个人体重x[i]，让你把他们分成两组，两组之间的人数差不能大于1。输出两组体重差最小的情况

Tug of War

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 7419		Accepted: 1956

Description

A tug of war is to be arranged at the local office picnic. For the tug of war, the picnickers must be divided into two teams. Each person must be on one team or the other; the number of people on the two teams must not differ by more than 1; the total weight of the people on each team should be as nearly equal as possible.

Input

The first line of input contains n the number of people at the picnic. n lines follow. The first line gives the weight of person 1; the second the weight of person 2; and so on. Each weight is an integer between 1 and 450. There are at most 100 people at the picnic.

Output

Your output will be a single line containing 2 numbers: the total weight of the people on one team, and the total weight of the people on the other team. If these numbers differ, give the lesser first.

Sample Input

Sample Output

190 200

#include<stdio.h>
#include<string.h>
int dp[45110][115];
int w[115];
int main()
{
    int i,j,k,l,m,n,sum,r;
   // scanf("%d",&m);
    while(scanf("%d",&n)!=EOF)
    {
        sum=0;
        r=(n+1)>>1;
        for(i=0; i<n; i++)
        {
            scanf("%d",&w[i]);
            sum+=w[i];
        }
        memset(dp,0,sizeof(dp));
        dp[0][0]=1;
        for(i=0; i<n; i++)
        {
            for(j=sum/2; j>=w[i]; j--)
            {
                for(k=r; k>=1; k--)
                {
                    if(dp[j-w[i]][k-1]==1)
                        dp[j][k]=1;
                }
            }
        }
        for(i=sum/2; i>=0; i--)
        {
            if(n%2==0)
            {
                if(dp[i][r]==1)
                    break;
            }
            else
            {
                if(dp[i][r]==1||dp[i][r-1]==1)
                    break;
            }
        }
        printf("%d %d\n",i,sum-i);
    }
    return 0;
}

/*
3
100
90
200
*/

poj 2576 简单 二维背包

poj 2576 简单二维背包