POJ 1700 cross river (mathematical simulation)

ZookeeperCrossing River

Time Limit:1000 MS		Memory Limit:10000 K
Total Submissions:10311		Accepted:3889

Description

A group of N people wishes to go into ss 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 started ss.

Input

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. each case is preceded by a blank line. there won't be more than 1000 people and nobody takes more than 100 seconds to cross.

Output

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

Sample Input

141 2 5 10

Sample Output

17

Train of Thought: (for example, 1-n people's time [n], in ascending order) when there is only one person: sum = time [1]; when there are two people: when sum = time [1] + time [2] 3: sum = time [1] + time [2] + time [3] Important! When there are more than four people, there are only two shipping methods in pursuit of the shortest time: (1) the fastest, the second is the fastest --> the fastest-> the slowest, slow trip --> fast time [2] + time [1] + time [n] + time [n-1] + time [2] (2) fastest, slowest --> fastest, fast --> fastest return time [n] + time [1] + time [n-1] + time [1]

#include
 
  #include#define maxn 100001using namespace std;int time[maxn];int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,sum=0;        scanf("%d",&n);        for(int i=1;i<=n;i++)            scanf("%d",time+i);        sort(time+1,time+n+1);        while(n)        {            if(n==1)            {                sum+=time[1];                n=0;            }            else if(n==2)            {                sum+=time[2];                n=0;            }            else if(n==3)            {                sum+=time[1]+time[2]+time[3];                n=0;            }            else            {                if(time[2]*2>=time[1]+time[n-1])                    sum+=2*time[1]+time[n]+time[n-1];                else                    sum+=2*time[2]+time[1]+time[n];                n-=2;            }        }        printf("%d\n",sum);    }    return 0;}