Crossing River（贪心+模拟）

Crossing River（贪心+模拟）

Crossing River

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit  ​​Status​​​  ​​​Practice​​​ ​​​POJ 1700​​

Description

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.

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

1 4 1 2 5 10

Sample Output

17

如果有N个人（N>=1） 如果N=1或N=2；所有人直接过河。 如果N=3；由最快的人往返一次把其他两人送过河。 如果N>=4;设A，B为走的最快的和次快的人，过桥时间为a，b； 设Z，Y是走的最慢和次慢的人，过桥时间为z,y： 当2b>a+y时，使用模式一将Z，Y送过桥 当2b

//过河问题#include#include#include#includeusing namespace std;int a[1005];int main(){ int t,n; scanf("%d",&t); while(t--) { scanf("%d",&n); for(int i=0;i3) { if(a[0]+a[n-1]+a[0]+a[n-2]>=a[0]+a[1]+a[n-1]+a[1]) sum+=a[0]+a[1]+a[n-1]+a[1]; else sum+=a[0]+a[n-1]+a[0]+a[n-2]; n-=2; } if(n==3) sum+=(a[0]+a[2]); printf("%d\n",sum); }}

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