[ZH模拟赛20171029T1]Cards

Problem

每张牌有两个数字，你可以选择使用哪个数字，然后你还可以任意排序，求一个加一个减后的最小值。

Solution

显然，我们可以先全部加上，然后再选择n/2个改为减去。
因此我们对每张牌取最小值先加上，然后把两个数字的和放入数组中进行排序，最后减去大的那n/2个就好了。

Notice

没啥难点

Code

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define sqz main
#define ll long long
#define reg register int
#define rep(i, a, b) for (reg i = a; i <= b; i++)
#define per(i, a, b) for (reg i = a; i >= b; i--)
#define travel(i, u) for (reg i = head[u]; i; i = edge[i].next)
const int INF = 1e9, N = 100000;
const double eps = 1e-6, phi = acos(-1.0);
ll mod(ll a, ll b) {if (a >= b || a < 0) a %= b; if (a < 0) a += b; return a;}
ll read(){ ll x = 0; int zf = 1; char ch; while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;}
void write(ll y) { if (y < 0) putchar('-'), y = -y; if (y > 9) write(y / 10); putchar(y % 10 + '0');}
int T[N + 5];
int sqz()
{
	int n = read(), sum = 0;
	rep(i, 1, n)
	{
		int a = read(), b = read();
		T[i] = a + b;
		sum += min(a, b);
	}
	sort(T + 1, T + n + 1);
	rep(i, 1, n / 2) sum -= T[n - i + 1];
	printf("%d\n", sum);
}