Candy

qwqq今天是平安夜呢

然而23道贪心我才做了一半多一点。。。

702:Crossing River

描述

A group of N people wishes to go across a river with only one boat, which can at most carry two persons. Therefore some sort of shuttle arrangement must be arranged in order to row the boat back and forth so that all people may cross. Each person has a different rowing speed; the speed of a couple is determined by the speed of the slower one. Your job is to determine a strategy that minimizes the time for these people to get across.

输入

The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. The first line of each case contains N, and the second line contains N integers giving the time for each people to cross the river. There won't be more than 1000 people and nobody takes more than 100 seconds to cross.

输出

For each test case, print a line containing the total number of seconds required for all the N people to cross the river.

其实这道题刚开始和綦天菲一起想测试数据是怎么算的...两人卒在电脑前（捂脸

以为后来我俩就这样想出来了么？

当然是没有啊。。。

偶然发现别人有一个很好的思路...简单易懂（可我就没想出来）

以下：

一些人想要过河，例如4个人过河的所需时间分别为：1，2，5，10
那么过河有两种方法：
1. 1与2，1回来，1与5，1回来，1与10  结果耗时19
2. 1与2，2回来，5与10，1回来，1与2  结果耗时17
设 x，y为最快和次快，a，b为次慢和最慢
那么第一种耗时t=y+x+a+x+b
第二种耗时t=y+y+b+x+y
重点就在于2*y和x+a谁大谁小
根据贪心的思想，每一次都比较一下呗
谁小选谁。

amazing

#include<iostream>
#include<algorithm>
using namespace std;
int a[1010];
int main()
{
int T,n,sum;
cin>>T;
while(T--){ 
cin>>n; 
for(int k=0;k<n;k++){
cin>>a[k];
} 
sum=0; 
sort(a,a+n);
if(n==1)
sum+=a[0];
while(n>1)
{
if(n==2)
sum+=a[1];
if(n==3)
sum+=(a[0]+a[1]+a[2]);
if(n>=4)
{
if(2*a[1]-a[n-2]-a[0]<0)
{
sum+=(a[0]+2*a[1]+a[n-1]);
}
else
{
sum+=(2*a[0]+a[n-2]+a[n-1]);
}
}
n-=2;
}
cout<<sum<<endl;
}

return 0;
};

曾以为走不出的日子，现在都回不去了。