Description

题库链接

双方 \(n\) 人，给出每人的战斗力，赢一场加 \(2\) 分，平局 \(1\) 分，失败不得分。求最大和最小的得分。

\(1\leq n\leq 100000\)

Solution

田忌赛马。贪心策略及证明可以参考链接

最小得分就相当于 \(2n-\) 对方最大得分。交换再做一次贪心即可。

Code

//It is made by Awson on 2018.2.27

#include <bits/stdc++.h>

#define LL long long

#define dob complex<double>

#define Abs(a) ((a) < 0 ? (-(a)) : (a))

#define Max(a, b) ((a) > (b) ? (a) : (b))

#define Min(a, b) ((a) < (b) ? (a) : (b))

#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))

#define writeln(x) (write(x), putchar('\n'))

#define lowbit(x) ((x)&(-(x)))

using namespace std;

const int N = 100000;

void read(int &x) {

    char ch; bool flag = 0;

    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());

    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());

    x *= 1-2*flag;

}

void print(int x) {if (x > 9) print(x/10); putchar(x%10+48); }

void write(int x) {if (x < 0) putchar('-'); print(Abs(x)); }

int n, a[N+5], b[N+5];

bool comp(const int &a, const int &b) {return a > b; }

int cal(int* a, int* b) {

    int l1 = 1, l2 = 1, r1 = n, r2 = n, ans = 0;

    while (l1 <= r1) {

    if (a[r1] > b[r2]) ans += 2, --r1, --r2;

    else if (a[l1] > b[l2]) ans += 2, ++l1, ++l2;

    else ans += (a[r1] == b[l2]), --r1, ++l2;

    }

    return ans;

}

void work() {

    read(n);

    for (int i = 1; i <= n; i++) read(a[i]); sort(a+1, a+n+1, comp);

    for (int i = 1; i <= n; i++) read(b[i]); sort(b+1, b+n+1, comp);

    write(cal(a, b)), putchar(' '), writeln(2*n-cal(b, a));

}

int main() {

    work(); return 0;

}