poj_3404_Bridge_over_a_rough_river.cpp

Problem Links:

poj3404,

Problem:

Bridge over a rough river

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3063		Accepted: 1211

Description

A group of N travelers (1 ≤ N ≤ 50) has approached an old and shabby bridge and wishes to cross the river as soon as possible. However, there can be no more than two persons on the bridge at a time. Besides it's necessary to light the way with a torch for safe crossing but the group has only one torch.

Each traveler needs t_i seconds to cross the river on the bridge; i=1, ... , N (t_i are integers from 1 to 100). If two travelers are crossing together their crossing time is the time of the slowest traveler.

The task is to determine minimal crossing time for the whole group.

Input

The input consists of two lines: the first line contains the value of N and the second one contains the values of t_i (separated by one or several spaces).

Output

The output contains one line with the result.

Sample Input

4
6 7 6 5

Sample Output

29

Source

Northeastern Europe 2001, Western Subregion

Solution:

Just like POJ 1700;

Source Code:

[sourcecode language="cpp" collapse="true" padlinenumbers="true"]
//Fri May 21 15:37:01 CDT 2010
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

long long crossriver(vector<int> v, int T)
{
if (T == 1)
{
return v[0];
}
if (T == 2)
{
return v[1];
}
if (T == 3)
{
return v[0] + v[1] + v[2];
}
if (v[0] + 3 * v[1] + v[T - 1] > 2 * v[0] + v[1] + v[T - 2] + v[T - 1])
{
return 2 * v[0] + v[T - 2] + v[T - 1] + crossriver(v, T - 2);
}
else
{
return v[0] + 2 * v[1] + v[T - 1] + crossriver(v, T - 2);
}
}

int main(int argc, const char* argv[])
{
// freopen("input.in", "r", stdin);
// freopen("output.out", "w", stdout);
int T;
while(cin >> T)
{
vector<int> v;
for(int i=0; i<T; i++)
{
int number;
cin >> number;
v.push_back(number);
}
sort(v.begin(), v.end());
cout << crossriver(v, T) << endl;
}
// fclose(stdin);
// fclose(stdout);
return 0;
}
[/sourcecode]