HDU 5461

Given the sequence AA with nn integers t1,t2,⋯,tnt1,t2,⋯,tn. Given the integral coefficients aaand bb. The fact that select two elements titi and tjtj of AA and i≠ji≠j to maximize the value of at2i+btjati2+btj, becomes the largest point.

InputAn positive integer TT, indicating there are TT test cases. 
For each test case, the first line contains three integers corresponding to n (2≤n≤5×106), a (0≤|a|≤106)n (2≤n≤5×106), a (0≤|a|≤106) and b (0≤|b|≤106)b (0≤|b|≤106). The second line contains nnintegers t1,t2,⋯,tnt1,t2,⋯,tn where 0≤|ti|≤1060≤|ti|≤106 for 1≤i≤n1≤i≤n. 

The sum of nn for all cases would not be larger than 5×1065×106.OutputThe output contains exactly TT lines. 
For each test case, you should output the maximum value of at2i+btjati2+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

只需要维护一下a*t*t和b*t各自的最大和次大值。然后记录最大值对应的下标，以此来判断冲突。
当时没有想清楚，以至于维护了太多最值，把自己搞乱了，也和最近身体状态有关吧。

#include<cstdio>
#include<algorithm>
using namespace std;
#define ll long long
const ll inf = 1000000000000000000ll;
int main(){
    int t;
    scanf("%d",&t);
    for(int i=1;i<=t;i++){
        ll n, a, b, t;
        ll max1a=-inf,max2a=-inf,max1b=-inf,max2b=-inf;
        int x=-1, y=-1;
        scanf("%lld%lld%lld",&n,&a,&b);
        for(int i=0;i<n;i++){
            scanf("%lld",&t);
            if(a*t*t>max1a){
                x = i;
                max2a = max1a;
                max1a = a*t*t;
            }else if(a*t*t>max2a){
                max2a = a*t*t;
            }
            if(b*t>max1b){
                y = i;
                max2b = max1b;
                max1b = b*t;
            }else if(b*t>max2b){
                max2b = b*t;
            }
        }
        printf("Case #%d: ",i);
        if(x==y){
            printf("%lld\n",max(max1a+max2b,max2a+max2b));
        }else
            printf("%lld\n",max1a+max1b);
    }
    return 0;
}