本文主要是介绍POJ1700 Crossing River 过河问题，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
一、题目信息
Crossing River
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 8897 | Accepted: 3323 |
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
二、参考代码
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;int main()
{int N;cin >> N;while(N--){vector<int> a;int n,m,sum = 0;cin >> n;for(int i = 1;i <= n;i++){cin >> m;a.push_back(m);}sort(a.begin(),a.end());a.insert(a.begin(),0);while(1){if(n == 2) {sum += a[2];break;}else if(n == 1){sum += a[1];break;}else if(n == 3){sum +=a[1] + a[2] + a[3];break;}else{sum += min(2*a[2] + a[1] + a[n] ,a[n] + 2*a[1] + a[n-1]);a.erase(a.end()-1,a.end());n -= 2;}}cout << sum << endl;}return 0;
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