BestCoder Round #72 (div.2) B.Clarke and points

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& hdoj 5626

题意：

平面上n个点，求两点间最大的曼哈顿距离。

题解：

假设A, B两点使得曼哈顿距离最大，去绝对值，可以化简如下：

则只要分别求出最大和最小的、即可。

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 1000000 + 10;struct point{    ll x, y;}a[maxn];ll seed;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(){    int t, n;    cin >> t;    while(t--)    {        cin >> n >> seed;        for (int i = 0; i < n; i++){            a[i].x = Rand(-1000000000, 1000000000),            a[i].y = Rand(-1000000000, 1000000000);        }        ll xMax = a[0].x + a[0].y, xMin = a[0].x + a[0].y;        ll yMax = a[0].x - a[0].x, yMin = a[0].x - a[0].y;        for(int i = 1; i < n; ++i)        {            ll ax = a[i].x + a[i].y;            ll in = a[i].x - a[i].y;            xMax = max(xMax, ax);            xMin = min(xMin, ax);            yMax = max(yMax, in);            yMin = min(yMin, in);        }        ll ans = max(abs(xMax - xMin), abs(yMax - yMin));        printf("%I64d\n", ans);    }    return 0;}

还有一种通过二进制运算来枚举这四种形式的，妙。

这种方法也很适合计算n维空间两点的曼哈顿距离。

参考点击打开链接

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 1000000 + 10;const ll inf = 1LL << 60;const int dimension = 2;long long seed;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);}struct point{    ll x[dimension];}p[maxn];int main(){    int t, n;    cin >> t;    while(t--)    {        cin >> n >> seed;        for (int i = 0; i < n; i++){            p[i].x[0] = Rand(-1000000000, 1000000000),            p[i].x[1] = Rand(-1000000000, 1000000000);        }        ll ans = 0;        for(int s = 0; s < (1 << dimension); ++s)        {            ll ax = -inf, in = inf;            for(int i = 0; i < n; ++i)            {                ll tmp = 0;                for(int j = 0; j < dimension; ++j)                {                    if((1 << j) & s) tmp += p[i].x[j];                    else tmp -= p[i].x[j];                }                ax = max(ax, tmp);                in = min(in, tmp);            }            ans = max(ax - in, ans);        }        cout << ans << endl;    }}

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