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2020牛客多校第3场：Two Matchings[找规律+dp]

2020年07月23日  | 移动技术网移动技术  | 我要评论

题目链接

解题思路：这题规律。。。。。。无语：
在这里插入图片描述

#include <iostream>
#include <cstdio>
#include <stack>
#include <sstream>
#include <vector>
#include <map>
#include <cstring>
#include <deque>
#include <cmath>
#include <iomanip>
#include <queue>
#include <algorithm>
#include <set>
#define mid ((l + r) >> 1)
#define Lson rt << 1, l , mid
#define Rson rt << 1|1, mid + 1, r
#define ms(a,al) memset(a,al,sizeof(a))
#define log2(a) log(a)/log(2)
#define _for(i,a,b) for( int i = (a); i < (b); ++i)
#define _rep(i,a,b) for( int i = (a); i <= (b); ++i)
#define for_(i,a,b) for( int i = (a); i >= (b); -- i)
#define rep_(i,a,b) for( int i = (a); i > (b); -- i)
#define lowbit(x) ((-x) & x)
#define IOS std::ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define INF 0x3f3f3f3f
#define hash Hash
#define next Next
#define count Count
#define pb push_back
#define f first
#define s second
using namespace std;
const int N = 2e5+10, mod = 1e9 + 7;
const long double eps = 1e-5;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> PLL;
typedef pair<double,double> PDD;
template<typename T> void read(T &x)
{
    x = 0;char ch = getchar();ll f = 1;
    while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
    while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
template<typename T, typename... Args> void read(T &first, Args& ... args)
{
    read(first);
    read(args...);
}
int n;
ll a[N];
ll dp[N];
int main()
{
    int T;
    read(T);
    while(T --)
    {
      read(n);
      for(int i = 1; i <= n; ++ i) read(a[i]);
      sort(a+1,a+n+1);
      ll ans = 0;
      for(int i = 2; i <= n; i += 2)
        ans += a[i] - a[i - 1];
      dp[2] = 1ll << 60;
      dp[4] = a[4] + a[3] - a[2] - a[1];
      dp[6] = a[6] + a[5] - a[4] + a[3] - a[2] - a[1]; 
      for(int i = 8; i <= n; i += 2)
      dp[i] = min(dp[i - 4] + a[i] + a[i - 1] - a[i - 2] - a[i - 3],
      dp[i - 6] + a[i] + a[i - 1] - a[i - 2] + a[i - 3] - a[i - 4] - a[i - 5]);
      cout << ans + dp[n] << endl;
    }
    return 0;
}

本文地址：https://blog.csdn.net/weixin_45630722/article/details/107523517

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