UVaLive 6694 Toy Boxes (二分+想法)

题意：给出n个数，把n个数放在三个盒子里，每个盒子里的数绑在一起，要拿出来任何一个数的时候，所承担的重量是整个盒子的总重量，求最小总重量和。

析：感觉吧，就是轻的放的多一些，拿的次数多一些，大的放的少一些，拿的少一些。分成两堆时一定是连续小的物品在一堆，连续大的在一堆，

也就是其中一堆的任意一个物品重量都要小于另一堆。,并且小的那一堆的个数一定要大于大的那一堆的个数。 

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <functional>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <deque>
#include <map>
#include <cctype>
#include <stack>
#include <sstream>
#include <cstdlib>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2e4 + 5;
const int mod = 1e9 + 7;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, -1, 1, 1, -1};
int n, m;
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
int a[maxn];
LL sum[maxn];
bool cmp(const int &lhs, const int &rhs){
    return lhs > rhs;
}

int main(){
    int T;    cin >> T;
    while(T--){
        scanf("%d", &n);
        for(int i = 1; i <= n; ++i)  scanf("%d", &a[i]);
        sort(a+1, a+n+1, cmp);
        sum[1] = a[1];
        for(int i = 2; i <= n; ++i)  sum[i] = a[i] + sum[i-1];
        LL ans = LNF;
        for(int i = 1; i <= n/3; ++i){
            for(int j = i+1; j <= n*2/3; ++j){
                int k = n - j;
                ans = Min(ans, sum[i]*(i) + (sum[j]-sum[i])*(j-i) + (sum[n]-sum[j]) * k);
            }
        }

        printf("%lld\n", ans);
    }
    return 0;
}