POJ 2573 Bridge 贪心

Bridge

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3220		Accepted: 1144		Special Judge

Description

n people wish to cross a bridge at night. A group of at most two people may cross at any time, and each group must have a flashlight. Only one flashlight is available among the n people, so some sort of shuttle arrangement must be arranged in order to return
the flashlight so that more people may cross.

Each person has a different crossing speed; the speed of a group is determined by the speed of the slower member. Your job is to determine a strategy that gets all n people across the bridge in the minimum time.

Input

The first line of input contains n, followed by n lines giving the crossing times for each of the people. There are not more than 1000 people and nobody takes more than 100 seconds to cross the bridge.
Output

The first line of output must contain the total number of seconds required for all n people to cross the bridge. The following lines give a strategy for achieving this time. Each line contains either one or two integers, indicating which person or people form
the next group to cross. (Each person is indicated by the crossing time specified in the input. Although many people may have the same crossing time the ambiguity is of no consequence.) Note that the crossings alternate directions, as it is necessary to return
the flashlight so that more may cross. If more than one strategy yields the minimal time, any one will do.
Sample Input

4
1
2
5
10

Sample Output

17
1 2
1
5 10
2
1 2

Source

Waterloo local 2000.09.30

百度下才发现是很经典的贪心问题……看来真是刷题太少

POJ 1700,2573,3404 都是一个类型，买一赠二好实惠啊……

#include <iostream>
#include <algorithm>
using namespace std;

int a[1010],n,i,nn,ans=0;
int main()
{
cin>>n;
for (i=1;i<=n;++i)
cin>>a[i];
if (n>1){
sort(a+1,a+n+1);
for (nn=n;nn>3;nn-=2)
ans+=a[1]+a[nn-1]<2*a[2]?2*a[1]+a[nn-1]+a[nn]:a[1]+2*a[2]+a[nn];
if (nn==2) ans+=a[2];
else ans+=a[1]+a[2]+a[3];
cout<<ans<<endl;

for (nn=n;nn>3;nn-=2)
if (a[1]+a[nn-1]<2*a[2])
cout<<a[1]<<' '<<a[nn-1]<<endl<<a[1]<<endl<<a[1]<<' '<<a[nn]<<endl<<a[1]<<endl;
else
cout<<a[1]<<' '<<a[2]<<endl<<a[1]<<endl<<a[nn-1]<<' '<<a[nn]<<endl<<a[2]<<endl;
if (nn==2)
cout<<a[1]<<' '<<a[2]<<endl;
else
cout<<a[1]<<' '<<a[2]<<endl<<a[1]<<endl<<a[1]<<' '<<a[3]<<endl;
}
if (n==1)
cout<<a[1]<<endl<<a[1]<<endl;

return 0;
}

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