HDU5461 Largest Point 思维 2015沈阳icpc

Largest Point

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 3065    Accepted Submission(s): 1078

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

 

 

题意：

　　求a*ti*ti+b*tj的最大值，ti，tj是num数组里的两个数

分析：

　　分情况考虑a和b为正为负的情况，当a，b为正时ti，tj要尽量大，因为a*ti*ti的影响大于b*tj，所以ti优先取最大值，在ti取完最大值后tj再来取剩下的最大值（这里用个map判断是否还剩余最大值，最大值可能有多个或者ti取了负的最大值）

　　然后是为负的情况用同样的方法计算就行

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
#define inf 0x17a6e6736e9
using namespace std;
const int maxn = 5*1e6 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
ll s1[maxn], s2[maxn];
int main() {
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    ll T, t = 0;
    cin >> T;
    while( T -- ) {
        t ++;
        ll n, a, b;
        cin >> n >> a >> b;
        map<ll,ll> mp;
        for( ll i = 0;  i < n; i ++ ) {
            cin >> s1[i];
            s2[i] = abs(s1[i]);
            mp[s1[i]] ++;
        }
        sort( s2, s2+n );
        sort( s1, s1+n );
        ll sum = 0;
        if( a > 0 ) {
            sum += a*s2[n-1]*s2[n-1];
            if( b > 0 ) {
                if( mp[-s2[n-1]] ) {
                    mp[-s2[n-1]] --;
                } else {
                    mp[s2[n-1]] --;
                }
                if( mp[s1[n-1]] > 0 ) {
                    sum += b*s1[n-1];
                } else {
                    sum += b*s1[n-2];
                }
            } else if( b < 0 ) {
                if( mp[s2[n-1]] ) {
                    mp[s2[n-1]] --;
                } else {
                    mp[-s2[n-1]] --;
                }
                if( mp[s1[0]] > 0 ) {
                    sum += b*s1[0];
                } else {
                    sum += b*s1[1];
                }
            }
        } else if( a < 0 ) {
            sum += a*s2[0]*s2[0];
            if( b > 0 ) {
                if( mp[-s2[0]] ) {
                    mp[-s2[0]] --;
                } else {
                    mp[s2[0]] --;
                }
                if( mp[s1[n-1]] > 0 ) {
                    sum += b*s1[n-1];
                } else {
                    sum += b*s1[n-2];
                }
            } else if( b < 0 ) {
                if( mp[s2[0]] ) {
                    mp[s2[0]] --;
                } else {
                    mp[-s2[0]] --;
                }
                if( mp[s1[0]] > 0 ) {
                    sum += b*s1[0];
                } else {
                    sum += b*s1[1];
                }
            }
        }
        cout << "Case #" << t << ": " << sum << endl;
    }
    return 0;
}