Light oj 1147 - Tug of War（状压dp）

1147 - Tug of War

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Time Limit: 4 second(s)

Memory Limit: 32 MB

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

Input starts with an integer T (≤ 100), denoting the number of test cases.

The first line of each case is a blank line. The next line of input contains an integer n (2 ≤ n ≤ 100), 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 100000. The summation of all the weights of the people in a case will not exceed 100000.

Output

For each case, print the case number and the total number weights of the people in two teams. If the weights differ, print the smaller weight first.

Sample Input	Output for Sample Input
2   3 100 90 200   4 10 15 17 20	Case 1: 190 200 Case 2: 30 32

 

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SPECIAL THANKS: JANE ALAM JAN (SOLUTION, DATASET)

1147 - Tug of War

/*

题意：有n个人，每个人都由重量。现将其分成人数差小于等于1的两组，求两组各自总质量之差最小的组合。
题解：动态规划，dp[i]=m,其中i为重量，m为二进制，
其第k位为1表示可以由 k个人组合使得总质量为i。之后就是01背包了。
//上面一句是重点

*/

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define bug printf("hihi\n")

#define eps 1e-8
typedef long long ll;

using namespace std;

#define INF 0x3f3f3f3f
#define N 100005

ll dp[N];
int a[105];
int all;
int n;

bool judge(int m)
{
    ll ha=dp[m];
    int t=n/2;
    if(n&1)
        return (ha&(1ll<<t)||(ha&(1ll<<(t+1))));
    return (ha&(1ll<<t));
}

int main()
{
    int i,j,t,ca=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        all=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            all+=a[i];
        }
        memset(dp,0,sizeof(dp));
        int m=all/2;
        dp[0]=1;
        for(i=1;i<=n;i++)
        {
             for(int v=m;v>=a[i];v--)
                  dp[v]=dp[v]|((ll)dp[v-a[i]]<<1);
        }

        for(i=m;i>=0;i--)
            if(judge(i))
        {
              printf("Case %d: %d %d\n",++ca,i,all-i);
              break;
        }
    }
    return 0;
}