Problem F: Tug of War UVA 10032 DP

G - Tug of War

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Submit Status

Appoint description:  System Crawler  (2013-05-30)

Description

 

Problem F: Tug of War

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 input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

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

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

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

1

3
100
90
200

Sample Output

190 200
T和WA了那叫一个惨！

/*
 * Author:  
 * Created Time:  2013/10/14 22:43:39
 * File Name: A.cpp
 * solve: A.cpp
 */
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<string>
#include<map>
#include<stack>
#include<set>
#include<iostream>
#include<vector>
#include<queue>
//ios_base::sync_with_stdio(false);
//#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;
#define sz(v) ((int)(v).size())
#define rep(i, a, b) for (int i = (a); i < (b); ++i)
#define repf(i, a, b) for (int i = (a); i <= (b); ++i)
#define repd(i, a, b) for (int i = (a); i >= (b); --i)
#define clr(x) memset(x,0,sizeof(x))
#define clrs( x , y ) memset(x,y,sizeof(x))
#define out(x) printf(#x" %d
", x)
#define sqr(x) ((x) * (x))
typedef long long LL;

const int INF = 1000000000;
const double eps = 1e-8;
const int maxn = 300;

int sgn(const double &x) {  return (x > eps) - (x < -eps); }
int num[maxn];
int dp[100][450*100 + 10];
int MIN[maxn];
int MAX[maxn];
bool cmp(int x ,int y)
{
    return x > y;
}
int main() 
{
    //freopen("in.txt","r",stdin);
    int T;
    scanf("%d",&T);
    int first = 0;
    while(T--)
    {
        if(first)
            printf("
");
        first = 1;
        int n;
        scanf("%d",&n);

        int sum = 0;
        repf(i,1,n)
        {
            scanf("%d",&num[i]);
            sum += num[i];
        }  

        clr(dp);

        dp[0][0] = 1;

        sort(num+1,num+1+n,cmp);

        MAX[0] = 0;
        repf(i,1,n)
            MAX[i] = MAX[i-1] + num[i];

        sort(num+1,num+1+n);

        MIN[0] = 0;
        repf(i,1,n)
            MIN[i] = MIN[i-1] + num[i];

        int N = n/2; 

        repf(i,1,n)
        {
            int temp = min(N,i);
            repd(k,temp - 1,0 )
                repd(j,MAX[k],MIN[k])
                {
                    if(dp[k][j])
                        dp[k+1][num[i] + j] = 1;
                }
        }


        int ans = INF;
        repd(i,MAX[N],0)
        {
            if(dp[N][i])
            {
                ans = min( ans , abs( sum - i - i ) );
            }
        }
        int m = ( sum - ans ) / 2;
        printf("%d %d
",m,sum-m);

    }
    return 0;
}

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