HDU 5461 Largest Point(2015沈阳赛区网络赛+技巧水题)

Largest Point

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 637 Accepted Submission(s): 266

Problem Description

Given the sequence A with n integers t1,t2,⋯,tn . Given the integral coefficients a and b . The fact that select two elements ti and tj of A and i≠j to maximize the value of at2i+btj , becomes the largest point.

Input

An positive integer T , indicating there are T test cases.
For each test case, the first line contains three integers corresponding to n (2≤n≤5×106), a (0≤|a|≤106) and b (0≤|b|≤106) . The second line contains n integers t1,t2,⋯,tn where 0≤|ti|≤106 for 1≤i≤n .

The sum of n for all cases would not be larger than 5×106 .

Output

The output contains exactly T lines.
For each test case, you should output the maximum value of at2i+btj .

Sample Input

   
   
   
   

    
    
    
    2

3 2 1
1 2 3

5 -1 0
-3 -3 0 3 3
   
   
   
   

Sample Output

   
   
   
   

    
    
    
    Case #1: 20
Case #2: 0
   
   
   
   

Source

2015 ACM/ICPC Asia Regional Shenyang Online 

/*小技巧：
分析：对于数组中的每一个t，我们用两个数组A和B分别纪录a*ti^2和b*ti，
然后对这两个数组排序，如果两个数组最大值的下标不同，那么相加就是最大值了，
如果相同，那么就拿A数组的次大值加上B数组的最大值，
和A数组的最大值加上B数字的次大值这两个值相比较，较大的即为最终的最大值。
*/

//改进不用快排 
#include <iostream>
#include <cstdio>
#include <string>
#include <string.h>
#include <algorithm>
#include <cstdlib>
#include <math.h>
using namespace std;
const int N=5e+6+10;
struct Node{
	long long v;
	int id;
}num1,num2,num11,num22;
int main(){
	int t,n,i;
	long long a,b,z;
	scanf("%d",&t);
	int now=1;
	while(t--){
          long long re=0;		
          num1.v=-(5e+18); 
           num11.v=-(5e+18); 
            num2.v=-(5e+18); 
             num22.v=-(5e+18); 
              num1.id=0; 
           num11.id=0; 
            num2.id=0; 
             num22.id=0; 
		scanf("%d%lld%lld",&n,&a,&b);
		for(i=0;i<n;i++){
			 scanf("%lld",&z);
		     if(a*z*z>num1.v){
		     	num11.v=num1.v;
		     	num11.id=num1.id;	
		     	num1.v=a*z*z;
		     	num1.id=i;	     	
			 }	
			 else if(a*z*z>num11.v){
			 	num11.v=a*z*z;
		     	num11.id=i;
			 }
			 if(b*z>num2.v){
		     	num22.v=num2.v;
		     	num22.id=num2.id;	
		     	num2.v=b*z;
		     	num2.id=i;	     	
			 }	
			 else if(b*z>num22.v){
			 	num22.v=b*z;
		     	num22.id=i;
			 }
		}	 
	 if(num1.id!=num2.id)
	     re=num1.v+num2.v;
	else
	  re=max(num1.v+num22.v,num11.v+num2.v);
	 printf("Case #%d: %lld\n",now++,re);
}
	return 0;
}



/*超时 
长得丑....... 
#include <iostream>
#include <cstdio>
#include <string>
#include <string.h>
#include <algorithm>
#include <cstdlib>
#include <math.h>
using namespace std;
const int N=5*1e6+10;
struct Node{
	long long v;
	int id;
	 bool operator <(const Node &b) const  {
	 if(v==b.v)
	  return id<b.id;
     return  v<b.v;
	 }
}num1[N],num2[N];
int main(){
	int t,n,i;
	long long a,b,z;
	scanf("%d",&t);
	int now=1;
	while(t--){
          long long re=0;		
		scanf("%d%lld%lld",&n,&a,&b);
		for(i=0;i<n;i++){
			 scanf("%lld",&z);
		    num1[i].id=i;
			num2[i].id=i;		
			 num1[i].v=a*z*z;
			 num2[i].v=b*z;			 
		}	 
		sort(num1,num1+n);
	    sort(num2,num2+n);
	 if(num1[n-1].id!=num2[n-1].id)
	     re=num1[n-1].v+num2[n-1].v;
	else
	  re=max(num1[n-1].v+num2[n-2].v,num1[n-2].v+num2[n-1].v);
	 printf("Case #%d: %lld\n",now++,re);
}
	return 0;
}

*/
/*
别人的代码951MS 过了 看脸啊.....日
#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=5e6+10;
int n,a,b;
struct node
{
    int id;
    long long x;
    bool operator <(const node &a) const
    {
        return x<a.x;
    }
}A[maxn],B[maxn];
int max(int x,int y)
{
    return x>y?x:y;
}
int main()
{
    int t,T=1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&a,&b);
        for(int i=0;i<n;i++)
        {
            long long v;
            scanf("%I64d",&v);
            A[i].x=a*v*v;
            B[i].x=b*v;
            A[i].id=B[i].id=i;
        }
        sort(A,A+n);
        sort(B,B+n);
        printf("Case #%d: ",T++);
        if(A[n-1].id!=B[n-1].id)
        {
            printf("%I64d\n",A[n-1].x+B[n-1].x);
            continue;
        }
        printf("%I64d\n",max(A[n-1].x+B[n-2].x,A[n-2].x+B[n-1].x));
    }
    return 0;
} 
*/