poj1700 Crossing River
阿新 • • 發佈：2017-08-15
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Crossing River
最大的和第二大的為n1,n2； 1、m2+m2<m1+n2；m1和m2先過河，然後m1回來，n1和n2過河。m2回來； 2、m2+m2>m1+n2。m1和n1先過河。然後m1回來。m1和n2過河。m1回來； 另外還要註意奇數的情況。
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 12585 | Accepted: 4787 |
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
Source
field=source&key=POJ+Monthly--2004.07.18" style="text-decoration:none">POJ Monthly--2004.07.18
題意：n個人一條船，全部人要過河；船每次僅僅能載兩個人。用時按最慢的算。求全部人過河的最小時間。
<span style="font-size:18px;">#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define MAXN 100010
int T,n;
int a[1010];
int sum;
int main()
{
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
CL(a, 0);
for(int i=0; i<n; i++)
scanf("%d",&a[i]);
if(n == 1){printf("%d\n",a[0]); continue;}
sort(a, a+n);
int m1 = a[0], m2 = a[1];
sum = 0;
for(int i=n-1; i>1; i-=2)
{
if(m2*2<m1+a[i-1])
{
if(i == 2) sum = sum+m1+m2+a[i];
else sum = sum+m1+m2*2+a[i];
}
else
{
sum = sum+m1*2+a[i]+a[i-1];
if(i == 2) sum -= a[0];
}
}
if(n!=0&&n%2==0) sum += m2;
printf("%d\n",sum);
}
return 0;
}
</span>最大的和第二大的為n1,n2； 1、m2+m2<m1+n2；m1和m2先過河，然後m1回來，n1和n2過河。m2回來； 2、m2+m2>m1+n2。m1和n1先過河。然後m1回來。m1和n2過河。m1回來； 另外還要註意奇數的情況。
poj1700 Crossing River