hdoj 5626 Clarke and points (曼哈顿距离)

最新推荐文章于 2022-08-05 21:35:57 发布

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最新推荐文章于 2022-08-05 21:35:57 发布

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本文介绍了一种高效计算给定点集间最大曼哈顿距离的方法，通过去除绝对值运算并采用四组线性组合的方式，实现了快速寻找最优解的目标。

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Clarke and points

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 433 Accepted Submission(s): 304

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

Sample Output

  
   1557439953
1423870062

Source

BestCoder Round #72 (div.2)

题意：按照题意求出每点的坐标，然后求出最大的曼哈顿距离

思路：嗯，题目数据挺大的，按部就班的算一定超时，

看了人家博客，思路妙，

思路就是去绝对值符号，然后就有四种情况：

(x1-y2)-(x2-y2); (x1+y1)-(x2+Y2); (-x1+y1)-(-x2+y2); (-x1-y1)-(-x2-y2). 减号左右类似，所以对每个（x,y）求出四种情况的值，然后求出最值

代码;

#include<cstdio>
#include<algorithm>
#define LL long long
using namespace std;

LL n,seed;
struct node
{
	LL x,y; 
}a[1000010];
LL rand(LL l, LL r) { 
  static LL mo=1e9+7, g=78125; 
  return l+((seed*=g)%=mo)%(r-l+1); 
} 
int main()
{
	int t;
	scanf("%d",&t);	
	while(t--)
	{
		scanf("%lld%lld",&n,&seed);
		for(int i=0;i<n;i++)
		{
			a[i].x = rand(-1000000000, 1000000000);
  			a[i].y = rand(-1000000000, 1000000000);
		}
		LL maxn[10],minn[10];
		for(int i=0;i<6;i++)
		{
			maxn[i]=-1e16;minn[i]=1e16;
		}
		for(int i=0;i<n;i++)
		{
			maxn[0]=max(maxn[0],a[i
			].x-a[i].y);
			maxn[1]=max(maxn[1],a[i].x+a[i].y);
			maxn[2]=max(maxn[2],-a[i].x+a[i].y);
			maxn[3]=max(maxn[3],-a[i].x-a[i].y);
			minn[0]=min(minn[0],a[i].x-a[i].y);
			minn[1]=min(minn[1],a[i].x+a[i].y);
			minn[2]=min(minn[2],-a[i].x+a[i].y);
			minn[3]=min(minn[3],-a[i].x-a[i].y);			
		}
		LL big=0;
		for(int i=0;i<4;i++)
		{
			big=max(big,maxn[i]-minn[i]);
		}
		printf("%lld\n",big);
	}
	return 0;
}