hdu 5626 Clarke and points

Problem Description

Clarke is a patient with multiple personality disorder. One day he turned into a learner of geometric.  He did a research on a interesting distance called Manhattan Distance. The Manhattan Distance between point A(xA,yA) and point B(xB,yB) is |xA−xB|+|yA−yB|.  Now he wants to find the maximum distance between two points of n points.

 

Input

The first line contains a integer T(1≤T≤5), the number of test case.  For each test case, a line followed, contains two integers n,seed(2≤n≤1000000,1≤seed≤109), denotes the number of points and a random seed.  The coordinate of each point is generated by the followed code. 
``` 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); }
// ...
cin >> n >> seed; for (int i = 0; i < n; i++)   x[i] = rand(-1000000000, 1000000000),   y[i] = rand(-1000000000, 1000000000); ```

 

Output

For each test case, print a line with an integer represented the maximum distance.

 

Sample Input

2 3 233 5 332

 

Sample Output

1557439953 1423870062

 

Source

BestCoder Round #72 (div.2)

 

先附上自己的写法，运气好的话可以过，运气不好的话超时，这东西也看人品？

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<vector>
#include<stdlib.h>
using namespace std;
#define ll long long
#define eps 1e-10
#define MOD 1000000007
#define N 1000006
#define inf 1e12

struct node{  
    ll x,y;  
}e[N],res[N];  
ll cmp(node a,node b)  
{  
    if(a.x==b.x)return a.y<b.y;  
    return a.x<b.x;  
}  
ll cross(node a,node b,node c)//向量积  
{  
    return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y);  
}  
ll convex(ll n)//求凸包上的点  
{  
    sort(e,e+n,cmp);  
    ll m=0,i,j,k;  
    //求得下凸包，逆时针  
    //已知凸包点m个，如果新加入点为i，则向量(m-2,i)必定要在(m-2,m-1)的逆时针方向才符合凸包的性质  
    //若不成立，则m-1点不在凸包上。  
    for(i=0;i<n;i++)  
    {  
        while(m>1&&cross(res[m-1],e[i],res[m-2])<=0)m--;  
        res[m++]=e[i];  
    }  
    k=m;  
    //求得上凸包  
    for(i=n-2;i>=0;i--)  
    {  
        while(m>k&&cross(res[m-1],e[i],res[m-2])<=0)m--;  
        res[m++]=e[i];  
    }  
    if(n>1)m--;//起始点重复。  
    return m;  
}  

long long n,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;
    scanf("%d",&t);
    while(t--){
        cin >> n >> seed;
        for (int i = 0; i < n; i++){
            e[i].x = rand(-1000000000, 1000000000),
            e[i].y = rand(-1000000000, 1000000000);
        }
        ll    m=convex(n);
        ll ans=-1;
        for(ll i=0;i<m;i++){
            for(ll j=i+1;j<m;j++){
                ll cnt = abs(res[i].x-res[j].x)+abs(res[i].y-res[j].y);
                ans=max(ans,cnt);
            }
        }
        printf("%I64d
",ans);
        
    }        
    return 0;
}

官方题解：

#include<bitset>
#include<map>
#include<vector>
#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<stack>
#include<queue>
#include<set>
#define inf 0x3f3f3f3f
#define mem(a,x) memset(a,x,sizeof(a))

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;

inline int in()
{
    int res=0;char c;int f=1;
    while((c=getchar())<'0' || c>'9')if(c=='-')f=-1;
    while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
    return res*f;
}
const int  N = 1000003;

ll a[N][3];
int n;
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;
    for (scanf("%d", &T);T--;) {
        cin >> n >> seed;
        for (int i=0; i<n; i++)
            a[i][0]=rand(-1000000000, 1000000000),
            a[i][1]=rand(-1000000000, 1000000000);
        ll t=0;
        ll ans=0,mx=-9223372036854775808LL,mn=9223372036854775807LL;
        for (int s=0; s<(1<<2); s++) {
            mx=-9223372036854775808LL,mn=9223372036854775807LL;
            for (int i=0; i<n; i++) {
                t = 0;
                for (int j=0; j<2; j++)
                    if ((1<<j) & s) t += a[i][j];
                    else t -= a[i][j];
                mn = min(mn, t);
                mx = max(mx, t);
            }
            ans = max(ans, mx-mn);
        }
        printf("%I64d
", ans);
    }
    return 0;
}