Codeforces 863B Kayaking 暴力 水题

传送门：http://codeforces.com/problemset/problem/863/B

啊啊啊啊啊……做的时候卡了好久啊，想多了，还觉得是dp啥的……其实就是个暴力……

暴力选两个人乘坐单人船，剩下的排序以后分为两组，记录差的最小值即可。

#include<iostream>
#include<string>
#include<string.h>
#include<cmath>
#include<algorithm>
#include<map>
using namespace std;
typedef long long ll;
int n,sum,ans;
int a[105],c[105];
int main()
{
    while (cin >> n)
    {
        sum = 0;
        for (int i = 0; i < 2 * n; i++)
        {
            cin >> a[i];
            sum += a[i];
        }
        sort(a, a + 2 * n);
        ans = 1e9;
        for (int i = 0; i < 2 * n; i++)
        {
            for (int j = i+1; j < 2 * n; j++)
            {
                int cur = 0,b=0;
                sum -= a[i];
                sum -= a[j];
                for (int k = 0; k < 2 * n; k++)
                {
                    if (k != i&&k != j)
                        c[cur++] = a[k];
                }
                for (int k = 0; k < 2 * n-2; k+=2)
                    b += c[k];
                ans = min(ans, abs(sum - b - b));
                sum += a[i] + a[j];
            }
        }
        cout << ans << endl;
    }
    return 0;
}

#include<iostream>
#include<string>
#include<string.h>
#include<cmath>
#include<algorithm>
#include<map>
using namespace std;
typedef long long ll;
int n,sum,ans;
int a[105],c[105];
int main()
{
    while (cin >> n)
    {
        sum = 0;
        for (int i = 0; i < 2 * n; i++)
        {
            cin >> a[i];
            sum += a[i];
        }
        sort(a, a + 2 * n);
        ans = 1e9;
        for (int i = 0; i < 2 * n; i++)
        {
            for (int j = i+1; j < 2 * n; j++)
            {
                int cur = 0,b=0;
                sum -= a[i];
                sum -= a[j];
                for (int k = 0; k < 2 * n; k++)
                {
                    if (k != i&&k != j)
                        c[cur++] = a[k];
                }
                for (int k = 0; k < 2 * n-2; k+=2)
                    b += c[k];
                ans = min(ans, abs(sum - b - b));
                sum += a[i] + a[j];
            }
        }
        cout << ans << endl;
    }
    return 0;
}