Codeforces Round #330 (Div. 2) C. Warrior and Archer（贪心博弈）

题意:

给定N≤2×105偶数个坐标,2个绝顶聪明的人玩游戏,轮流ban掉N−2个坐标
甲要使得最终2个尽可能近,乙则反之
求最终两个坐标的距离

分析:

假设[al,ar]是最终留下的坐标,可以发现甲一定选区间内的点,乙则反之
由于博弈,r−l≤n2,那么ans=minn2i=1{ai+n2−ai}

代码:

////  Created by TaoSama on 2016-01-21//  Copyright (c) 2015 TaoSama. All rights reserved.//#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 2e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, a[N];int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    while(scanf("%d", &n) == 1) {        for(int i = 1; i <= n; ++i) scanf("%d", a + i);        sort(a + 1, a + 1 + n);        int ans = INF;        for(int i = 1; i <= n >> 1; ++i)            ans = min(ans, a[i + n / 2] - a[i]);        printf("%d\n", ans);    }    return 0;}