浙江科技学院第十三届程序设计竞赛1008

A Heavy Rainy Day

Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 65535/32768K (Java/Other)
Total Submission(s) : 475   Accepted Submission(s) : 124

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Problem Description

As you know, Hangzhou always rains heavily, especially in Xiaoheshan area, which does annoy ACMers for a longer time. One day, N hardworking ACMers were
playing League of Legends in the lab, of course not include me, while it was raining cats and dogs. Everyone wanted to come back to the department where they were living together, but no one wanted to be soaked without umbrella. Unfortunately, they only had
one umbrella, which is too small to take more than two individuals. The time that everyone needs to spent to go to the department is known. If two guys came back together, it will take the longer time of this two guys. The question is the earliest time they
come back to the department.

Input

The first line contains an integer T(1<=T<=20),which is the number of test cases.

For each testcase:

The first line contains one integer N ,1<=N<=1000, present N ACMers want to come back,

The second line contains N integers Si, present the time the ith ACMers need.

Output

The shortest time they need to come to the department.

Sample Input

1
4
1 2 5 10

Sample Output

17

这道题需要四个一组考虑最优解

<span style="font-family:KaiTi_GB2312;font-size:24px;">#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
int a[1010];
int main()
{
int n;
scanf("%d",&n);
while(n--)
{
int m;
scanf("%d",&m);
for(int i=0;i<m;i++)
{
scanf("%d",&a[i]);
}
sort(a,a+m);
int sum=0,q,p;
while(m>3)
{
q=a[0]+a[1]*3+a[m-1];
p=a[0]*2+a[1]+a[m-1]+a[m-2];
if(q>p)
sum+=p-a[1];
else sum+=q-a[1];
m-=2;
}
if(m==1)
sum+=a[0];
if(m==2)
sum+=a[1];
if(m==3)
sum+=a[0]+a[1]+a[2];
printf("%d\n",sum);
}
return 0;
}</span>