HDU 5626 Clarke and points 求平面最远哈曼顿距离

来源：互联网发布：mac上好用的vpn 知乎编辑：程序博客网时间：2024/06/05 03:08

题意:根据题意给一个函数能得到一个点的数组，然后求所有点中最远的哈曼顿距离
思路:哈曼顿距离是|Xa-Xb|+|Ya-Yb|，这里看数据就知道不能裸跑，我们可以讨论绝对值内的正负，然后会出现四种情况，整理得到(Xa+Ya)-(Xb+Yb),(Xa-Ya)-(Xb-Yb),(-Xa+Ya)-(-Xb+Yb),(-Xa-Ya)-(-Xb-Yb)
(比赛的时候稍微算了一下，感觉nlogn能过，然后就T了QAQ
这里保存一下每个状态的最大最小就OK了

#include<map>#include<set>#include<queue>#include<stack>#include<math.h>#include<string>#include<vector>#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm> using namespace std;typedef long long LL;#define maxn 1000005#define f(x) (x*1.0)#define inf 0x7fffffff#define maxm maxn*maxn#define min(a,b) (a>b?b:a)#define max(a,b) (a>b?a:b)#define lowbit(x) (x&(-x))#define cheak(i) printf("%d ",i)#define lson(x) (splay[x].son[0])#define rson(x) (splay[x].son[1])#define rfor(i,a,b) for(i=a;i<=b;++i)#define lfor(i,a,b) for(i=a;i>=b;--i)#define mem(a,b) memset(a,b,sizeof(a))#define mec(a,b) memcpy(a,b,sizeof(b))struct node{    LL x,y;}A[maxn];LL B[maxn];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);}int main(){    int T,i;    LL n;    scanf("%d",&T);    while(T--)    {        scanf("%lld%lld",&n,&seed);        LL Min1=inf,Min2=inf,Min3=inf,Min4=inf,Max1=-inf,Max2=-inf,Max3=-inf,Max4=-inf;        rfor(i,1,n)        {            A[i].x = rand(-1000000000, 1000000000),            A[i].y = rand(-1000000000, 1000000000);            Min1=min(Min1,A[i].x+A[i].y);Max1=max(Max1,A[i].x+A[i].y);            Min2=min(Min2,A[i].x-A[i].y);Max2=max(Max2,A[i].x-A[i].y);            Min3=min(Min3,-A[i].x+A[i].y);Max3=max(Max3,-A[i].x+A[i].y);            Min4=min(Min4,-A[i].x-A[i].y);Max4=max(Max4,-A[i].x-A[i].y);            //printf("%lld %lld\n",A[i].x,A[i].y);        }        LL ans=0;        ans=max(ans,Max1-Min1);        ans=max(ans,Max2-Min2);        ans=max(ans,Max3-Min3);        ans=max(ans,Max4-Min4);         printf("%lld\n",ans);    }    return 0;}

0 0