hdu 5626 Clarke and points 最大曼哈顿距离

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题目：点击打开链接

题意：给n(<=10^6)个点,求两点的最远曼哈顿距离。

分析：这题和poj2926那题一样。

|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

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<cmath>#include<vector>#include<queue>#include<map>using namespace std;typedef long long ll;const int N=1000000+5;#define INF 20000000000;long long seed;int n;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);}ll f[N][2];int main(){   // freopen("f.txt","r",stdin);    ios_base::sync_with_stdio(0);   int T;   scanf("%d",&T);   while(T--){        cin >> n >> seed;        for (int i = 0; i < n; i++)            f[i][0] = rand(-1000000000, 1000000000),            f[i][1] = rand(-1000000000, 1000000000);        long long ans=0;         for (int k = 0; k < (1 << 2); k++){           long long a = -2000000000;            long long b = 2000000000;            for (int i = 0; i < n; i++)            {                ll temp = 0;                for (int j = 0; j < 2; j++)                    if(k>>j &1)temp+=f[i][j];                    else temp-=f[i][j];                a = max(a, temp);                b = min(b, temp);            }            ans = max(ans, a - b);        }        cout<<ans<<endl;   }    return 0;}

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