POJ 2576 二维背包

Tug of War

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 8437		Accepted: 2292

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

3
100
90
200

Sample Output

190 200

Source

Waterloo local 2000.09.30

题目意思：

    一个组有n个人，现在需要把这n个人分成两组，两组的人数的个数相差不超过1，并且使得二组的人的体重之和相差尽可能小；

解题思路：

   二位背包，nn=(n+1)>>1,sum=sum_weight>>1;在此基础上进行背包；

#include<iostream>
#include<cstdio>
#include<cstring>
#include<ctime>
#include<algorithm>
#include <map>
#include <queue>
using namespace std;
const int maxn=57;
const int maxs=20500;
bool dp[maxn][maxs];
int weight[107];
int n;
int sum;
int slove()
{
    memset(dp,0,sizeof(dp));
    dp[0][0]=1;
    int ans=0;
    int nn=(n+1)>>1;
    int sum2=sum>>1;
    for(int k=1;k<=n;k++)
    for(int i=nn;i>=1;i--)
        for(int j=sum2;j>=weight[k];j--)

    {
           if(!dp[i][j]) dp[i][j]=dp[i-1][j-weight[k]];
           if(dp[i][j]&&j>ans) ans=j;
    }

    return ans;
}
int main()
{
    while(~scanf("%d",&n)){
        sum=0;
        for(int i=1;i<=n;i++)
        {
          scanf("%d",&weight[i]);
          sum+=weight[i];
        }
        int ans=slove();
        int ans2=sum-ans;
        if(ans>ans2) swap(ans,ans2);
        printf("%d %d\n",ans,ans2);
    }
}