HDU 5626 Clarke and points（二进制枚举）

题意:

求N≤106个点中,任意2点间的最大曼哈顿距离

分析:

|xi−xj|+|yi−yj|,拆绝对值可以得到：
正正:xi−xj+yi−yj=(xi+yi)−(xj+yj)
负负:−xi+xj−yi+yj=(−xi−yi)−(−xj−yj)
正负:xi−xj−yi+yj=(xi−yi)−(xj−yj)
负正:−xi+xj+yi−yj=(−xi+yi)−(−xj+yj)
所以只要维护最大值和最小值就可以了,至于符号可以用二进制枚举
时间复杂度O(22∗2∗n)

代码:

////  Created by TaoSama on 2016-02-13//  Copyright (c) 2016 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 = 1e6 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;typedef long long LL;int n;long long seed, p[N][2];inline long long rand(long long l, long long r) {    static long long mo = 1e9 + 7, g = 78125;    return l + ((seed *= g) %= mo) % (r - l + 1);}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);    int t; scanf("%d", &t);    while(t--) {        scanf("%d%I64d", &n, &seed);        for(int i = 1; i <= n; ++i) {            p[i][0] = rand(-1000000000, 1000000000);            p[i][1] = rand(-1000000000, 1000000000);        }        LL ans = 0;        for(int s = 0; s < 1 << 2; ++s) {            LL minv = 1e18, maxv = -1e18;            for(int i = 1; i <= n; ++i) {                LL cur = 0;                for(int j = 0; j < 2; ++j)                    if(s >> j & 1) cur += p[i][j];                    else cur -= p[i][j];                minv = min(minv, cur);                maxv = max(maxv, cur);            }            ans = max(ans, maxv - minv);        }        printf("%I64d\n", ans);    }    return 0;}