标签：remove core code rds seq cal res output second
Let N be a positive integer.
There is a numerical sequence of length 3N, a=(a1,a2,…,a3N). Snuke is constructing a new sequence of length 2N, a‘, by removing exactly N elements from a without changing the order of the remaining elements. Here, the score of a‘ is defined as follows: (the sum of the elements in the first half of a‘)?(the sum of the elements in the second half of a‘).
Find the maximum possible score of a‘.
Input is given from Standard Input in the following format:
N a1 a2 … a3NOutput
Print the maximum possible score of a‘.
Sample Input 12 3 1 4 1 5 9Sample Output 1
1
When a2 and a6 are removed, a‘ will be (3,4,1,5), which has a score of (3+4)?(1+5)=1.
Sample Input 21 1 2 3Sample Output 2
-1
For example, when a1 are removed, a‘ will be (2,3), which has a score of 2?3=?1.
Sample Input 33 8 2 2 7 4 6 5 3 8Sample Output 3
5
For example, when a2, a3 and a9 are removed, a‘ will be (8,7,4,6,5,3), which has a score of (8+7+4)?(6+5+3)=5.
要求删去N个数之后，前面一半的sum-后面一半的sum最大值；
我们可以用一个小根堆维护前面，用一个大根堆维护后面；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 2000005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-4
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == ‘-‘) f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
int n;
ll a[maxn];
ll lftmax[maxn], minrgt[maxn];
int main() {
// ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
cin >> n;
for (int i = 0; i <3* n; i++)rdint(a[i]);
priority_queue<int, vector<int>, greater<int> >q1;
priority_queue<int>q2;
for (int i = 0; i < n; i++) {
lftmax[0] += a[i]; q1.push(a[i]);
minrgt[n + 1] += a[3 * n - i - 1]; q2.push(a[3 * n - i - 1]);
}
for (int i = 1; i <= n; i++) {
q1.push(a[n + i - 1]);
lftmax[i] = lftmax[i - 1] + a[n + i - 1] - q1.top();
q1.pop();
q2.push(a[n * 2 - i]);
minrgt[n - i + 1] = minrgt[n + 1 - i + 1] + a[n * 2 - i] - q2.top();
q2.pop();
}
ll res = -inf;
for (int i = 0; i <= n; i++) {
if (i == 0)res = lftmax[i] - minrgt[i + 1];
else res = max(res, lftmax[i] - minrgt[i + 1]);
}
cout << res << endl;
return 0;
}
标签：remove core code rds seq cal res output second
原文地址：https://www.cnblogs.com/zxyqzy/p/10319227.html
