Poj 3404 bridge over a rough River (Bridge problem)

Question link: http://poj.org/problem? Id = 3404

Bridge over a rough River

Time limit:1000 ms		Memory limit:65536 K
Total submissions:4024		Accepted:1644

Description

A groupNTravelers (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 needsTiSeconds to cross the river on the bridge;I= 1 ,...,N(TiAre 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 valueNAnd the second one contains the valuesTi(Separated by one or several spaces ).

Output

The output contains one line with the result.

Sample Input

46 7 6 5

Sample output

29

Source

Northeastern Europe 2001, western subregion

Ideas:

When N = 1, the answer is output directly.

When N = 2, the maximum output value is

When N = 3, output three and

When N> = 4, the two strategies are converted into four persons to find the optimal solution. If the speed of crossing a river is set to a <B <C <D, there are two strategies:

First AB, a, CD, and B, that is, temp = a + 2 * B + d

Ad first, a back, AC back, a back, that is, temp = 2 * A + C + D (forget this situation)

Then, take the minimum value. That is to say, select solution 1 when 2 * B <A + C; otherwise, select solution 2.

The Code is as follows:

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <cstdlib>#include <climits>#include <ctype.h>#include <queue>#include <stack>#include <vector>#include <deque>#include <set>#include <map>#include <iostream>#include <algorithm>using namespace std;#define PI acos(-1.0)#define INF 0x3fffffff//typedef long long LL;//typedef __int64 LL;const int M = 1017;int main(){int n;int a[M];int i;while(~scanf("%d",&n)){for(i = 1; i <= n; i++){scanf("%d",&a[i]);}if(n == 1){printf("%d\n",a[1]);continue;}sort(a+1,a+n+1);if(n == 2){printf("%d\n",a[2]);continue;}if(n == 3){printf("%d\n",a[1]+a[2]+a[3]);continue;}int sum = 0;while(n){if(2*a[2] < a[1]+a[n-1]){sum+=a[1]+2*a[2]+a[n];n-=2;}else{sum+=2*a[1]+a[n-1]+a[n];n-=2;}if(n <= 3)break;}if(n == 1)sum+=a[1];else if(n == 2)sum+=a[2];else if(n == 3){printf("%d\n",sum+a[1]+a[2]+a[3]);continue;}printf("%d\n",sum);}return 0;}